From 812e9bab668697e018453694eab480f084d2c75a Mon Sep 17 00:00:00 2001 From: Ainevsia Date: Fri, 1 Sep 2023 07:47:44 +0000 Subject: [PATCH] =?UTF-8?q?Deploying=20to=20gh-pages=20from=20@=20Ainevsia?= =?UTF-8?q?/Leetcode-Rust@cb020718e6969afeeb5f783ebac45f080b9f9025=20?= =?UTF-8?q?=F0=9F=9A=80?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 404.html | 2 +- day1.html | 2 +- day1/lc27.html | 2 +- day1/lc704.html | 2 +- day10.html | 2 +- day10/lc232.html | 2 +- day10/lc255.html | 2 +- day11.html | 2 +- day11/lc1047.html | 2 +- day11/lc150.html | 2 +- day11/lc20.html | 2 +- day12.html | 2 +- day13.html | 2 +- day13/lc239.html | 2 +- day13/lc347.html | 2 +- day14.html | 2 +- day14/lc144.html | 2 +- day15.html | 2 +- day15/lc101.html | 2 +- day15/lc102.html | 2 +- day15/lc226.html | 2 +- day16.html | 2 +- day16/lc104.html | 2 +- day16/lc111.html | 2 +- day16/lc222.html | 2 +- day17.html | 2 +- day17/lc110.html | 2 +- day17/lc257.html | 2 +- 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a/day36/lc763.html +++ b/day36/lc763.html @@ -83,7 +83,7 @@ diff --git a/day37.html b/day37.html index 8d3259c..27cca3c 100644 --- a/day37.html +++ b/day37.html @@ -83,7 +83,7 @@ diff --git a/day37/lc738.html b/day37/lc738.html index d4843dc..ea8fb40 100644 --- a/day37/lc738.html +++ b/day37/lc738.html @@ -83,7 +83,7 @@ diff --git a/day37/lc968.html b/day37/lc968.html index 5d015e8..03a5b11 100644 --- a/day37/lc968.html +++ b/day37/lc968.html @@ -83,7 +83,7 @@ diff --git a/day38.html b/day38.html index ef543a7..45e6c80 100644 --- a/day38.html +++ b/day38.html @@ -83,7 +83,7 @@ diff --git a/day38/lc509.html b/day38/lc509.html index 87e94b7..43510f5 100644 --- a/day38/lc509.html +++ b/day38/lc509.html @@ -83,7 +83,7 @@ diff --git a/day38/lc70.html b/day38/lc70.html index 7198905..2b27a78 100644 --- a/day38/lc70.html +++ b/day38/lc70.html @@ -83,7 +83,7 @@ diff --git a/day38/lc746.html b/day38/lc746.html index 3626d52..3921664 100644 --- a/day38/lc746.html +++ b/day38/lc746.html @@ -83,7 +83,7 @@ diff --git a/day39.html b/day39.html index 3807f72..0992f82 100644 --- a/day39.html +++ b/day39.html @@ -83,7 +83,7 @@ diff --git a/day39/lc62.html b/day39/lc62.html index beb0472..edb68fd 100644 --- a/day39/lc62.html +++ b/day39/lc62.html @@ -83,7 +83,7 @@ diff --git a/day39/lc63.html b/day39/lc63.html index 09345d6..d43f290 100644 --- a/day39/lc63.html +++ b/day39/lc63.html @@ -83,7 +83,7 @@ diff --git a/day4.html b/day4.html index dbb527f..0b8b8f9 100644 --- a/day4.html +++ b/day4.html @@ -83,7 +83,7 @@ diff --git a/day4/lc02.07.html b/day4/lc02.07.html index 5dfa80f..fbcb966 100644 --- a/day4/lc02.07.html +++ b/day4/lc02.07.html @@ -83,7 +83,7 @@ diff --git a/day4/lc142.html b/day4/lc142.html index e4669b9..d17dcf4 100644 --- a/day4/lc142.html +++ b/day4/lc142.html @@ -83,7 +83,7 @@ diff --git a/day4/lc19.html b/day4/lc19.html index 067017c..caadb5e 100644 --- a/day4/lc19.html +++ b/day4/lc19.html @@ -83,7 +83,7 @@ diff --git a/day4/lc24.html b/day4/lc24.html index 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b/day45/lc322.html index ae5c18c..557cc23 100644 --- a/day45/lc322.html +++ b/day45/lc322.html @@ -83,7 +83,7 @@ diff --git a/day45/lc70.html b/day45/lc70.html index 5b64e43..12db337 100644 --- a/day45/lc70.html +++ b/day45/lc70.html @@ -83,7 +83,7 @@ diff --git a/day46.html b/day46.html index f1c06d7..89c72bb 100644 --- a/day46.html +++ b/day46.html @@ -83,7 +83,7 @@ @@ -188,7 +188,7 @@

- @@ -202,7 +202,7 @@

- diff --git a/day46/lc139.html b/day46/lc139.html new file mode 100644 index 0000000..53ec02a --- /dev/null +++ b/day46/lc139.html @@ -0,0 +1,241 @@ + + + + + + 139.单词拆分 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

139. 单词拆分

+

不会

+
class Solution {
+public:
+    bool wordBreak(string s, vector<string>& wordDict) {
+        // dp[i] -> 长度为i的字符串是否可以拼出来
+        // dp[j] = true if dp[j-i] == true and [i:j] in wordDict
+        unordered_set<string> wordSet(wordDict.begin(),wordDict.end());
+        vector<bool>dp(s.size()+1,false);
+        dp[0]=true;
+        for (int i = 1 ; i<=s.size();i++) {
+            for (int j = 0; j < i; j ++ ) {
+                string word = s.substr(j,i-j);
+                if (dp[j] && wordSet.find(word)!= wordSet.end())
+                    dp[i] = true;
+            }
+        }
+        return dp[s.size()];
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day48.html b/day48.html new file mode 100644 index 0000000..6fa218f --- /dev/null +++ b/day48.html @@ -0,0 +1,235 @@ + + + + + + day 48 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

第九章 动态规划part09

+

● 198.打家劫舍 +● 213.打家劫舍II
+● 337.打家劫舍III

+

详细布置

+

今天就是打家劫舍的一天，这个系列不算难，大家可以一口气拿下。

+

198.打家劫舍

+

视频讲解：https://www.bilibili.com/video/BV1Te411N7SX +https://programmercarl.com/0198.%E6%89%93%E5%AE%B6%E5%8A%AB%E8%88%8D.html

+

213.打家劫舍II

+

视频讲解：https://www.bilibili.com/video/BV1oM411B7xq +https://programmercarl.com/0213.%E6%89%93%E5%AE%B6%E5%8A%AB%E8%88%8DII.html

+

337.打家劫舍III

+

视频讲解：https://www.bilibili.com/video/BV1H24y1Q7sY +https://programmercarl.com/0337.%E6%89%93%E5%AE%B6%E5%8A%AB%E8%88%8DIII.html

+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day48/lc198.html b/day48/lc198.html new file mode 100644 index 0000000..ecfdf7d --- /dev/null +++ b/day48/lc198.html @@ -0,0 +1,239 @@ + + + + + + 198.打家劫舍 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

198. 打家劫舍

+

这个是抄随想录的

+
class Solution {
+public:
+    int rob(vector<int>& nums) {
+        if (nums.size() == 0) return 0;
+        if (nums.size() == 1) return nums[0];
+        // dp[i]：考虑下标i（包括i）以内的房屋，最多可以偷窃的金额为dp[i]
+        // dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        vector<int> dp(nums.size());
+        dp[0] = nums[0];
+        dp[1] = max(nums[0], nums[1]);
+        for (int i = 2; i < nums.size(); i++) {
+            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+        return dp[nums.size() - 1];
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day48/lc213.html b/day48/lc213.html new file mode 100644 index 0000000..2eefd03 --- /dev/null +++ b/day48/lc213.html @@ -0,0 +1,250 @@ + + + + + + 213.打家劫舍II - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

213. 打家劫舍 II

+

这个是自己写的

+
class Solution {
+public:
+    int rob(vector<int>& nums) {
+        if (nums.size() == 0) return 0;
+        if (nums.size() == 1) return nums[0];
+        // dp[i]：考虑下标i（包括i）以内的房屋，最多可以偷窃的金额为dp[i]
+        // dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        vector<int> dp(nums.size());
+        // skip index 0 first
+        int res = 0;
+        dp[0] = 0;
+        dp[1] = nums[1];
+        for (int i = 2; i < nums.size(); i++) {
+            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+        res = dp[nums.size()-1];
+        dp[0] = nums[0];
+        dp[1] = max(nums[0], nums[1]);
+        
+        for (int i = 2; i < nums.size() - 1; i++) {
+            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+
+        res = max(res,dp[nums.size()-2]);
+        return res;
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day48/lc337.html b/day48/lc337.html new file mode 100644 index 0000000..6415400 --- /dev/null +++ b/day48/lc337.html @@ -0,0 +1,254 @@ + + + + + + 337.打家劫舍III - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

337.打家劫舍III

+

自己做的

+
/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+// <r1,r2> r1 如果偷当前节点，得到的最大值 ； r2 不偷当前节点的最大值
+    pair<int,int> f(TreeNode*root) {
+        if(!root){return make_pair(0,0);}
+        auto leftr = f(root->left);
+        auto rightr = f(root->right);
+        int takeleft = leftr.first;
+        int notakeleft = leftr.second;
+        int takeright = rightr.first;
+        int notakeright = rightr.second;
+        int r1 = notakeleft + notakeright + root->val;
+        int r2 = max(takeleft,notakeleft) + max(takeright,notakeright);
+        return make_pair(r1, r2);
+    }
+    int rob(TreeNode* root) {
+        auto res = f(root);
+        return max(res.first, res.second);
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day49.html b/day49.html new file mode 100644 index 0000000..f8f5566 --- /dev/null +++ b/day49.html @@ -0,0 +1,231 @@ + + + + + + day 49 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

第九章 动态规划part10

+

● 121. 买卖股票的最佳时机 +● 122.买卖股票的最佳时机II

+

详细布置

+

股票问题是一个动态规划的系列问题，今日安排的题目不多，大家可以慢慢消化。

+

121. 买卖股票的最佳时机

+

视频讲解：https://www.bilibili.com/video/BV1Xe4y1u77q +https://programmercarl.com/0121.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BA.html

+

122.买卖股票的最佳时机II

+

视频讲解：https://www.bilibili.com/video/BV1D24y1Q7Ls +https://programmercarl.com/0122.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAII%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html

+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day49/lc121.html b/day49/lc121.html new file mode 100644 index 0000000..961ab3d --- /dev/null +++ b/day49/lc121.html @@ -0,0 +1,237 @@ + + + + + + 121. 买卖股票的最佳时机 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

121. 买卖股票的最佳时机

+

一开始这样子没有用动规

+
class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int res = 0;
+        int cmin = INT_MAX;
+        for (int i = 0; i < prices.size(); i ++ ) {
+            cmin = min(cmin, prices[i]);
+            if (prices[i] > cmin) {
+                res = max(res, prices[i] - cmin);
+            }
+        }
+        return res;
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day49/lc122.html b/day49/lc122.html new file mode 100644 index 0000000..58ab5ec --- /dev/null +++ b/day49/lc122.html @@ -0,0 +1,252 @@ + + + + + + 122.买卖股票的最佳时机II - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

122. 买卖股票的最佳时机 II

+

抄了随想录

+
class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int len = prices.size();
+        vector<vector<int>> dp(len, vector<int>(2, 0));
+        // dp[i][0] 表示第i天持有股票所得现金。
+        // dp[i][1] 表示第i天不持有股票所得最多现金
+
+// 如果第i天持有股票即dp[i][0]， 那么可以由两个状态推出来
+
+// 第i-1天就持有股票，那么就保持现状，所得现金就是昨天持有股票的所得现金 即：dp[i - 1][0]
+// 第i天买入股票，所得现金就是昨天不持有股票的所得现金减去 今天的股票价格 即：dp[i - 1][1] - prices[i]
+
+
+// 再来看看如果第i天不持有股票即dp[i][1]的情况， 依然可以由两个状态推出来
+
+// 第i-1天就不持有股票，那么就保持现状，所得现金就是昨天不持有股票的所得现金 即：dp[i - 1][1]
+// 第i天卖出股票，所得现金就是按照今天股票价格卖出后所得现金即：prices[i] + dp[i - 1][0]
+        dp[0][0] -= prices[0];
+        dp[0][1] = 0;
+        for (int i = 1; i < len; i++) {
+            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]); // 注意这里是和121. 买卖股票的最佳时机唯一不同的地方。
+            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i]);
+        }
+        return dp[len - 1][1];
+
+
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day5.html b/day5.html index 40ac50f..33ebc38 100644 --- a/day5.html +++ b/day5.html @@ -83,7 +83,7 @@ diff --git a/day50.html b/day50.html new file mode 100644 index 0000000..4c8f265 --- /dev/null +++ b/day50.html @@ -0,0 +1,232 @@ + + + + + + day 50 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

第九章 动态规划part11

+

● 123.买卖股票的最佳时机III
+● 188.买卖股票的最佳时机IV

+

详细布置

+

123.买卖股票的最佳时机III

+

这道题一下子就难度上来了，关键在于至多买卖两次，这意味着可以买卖一次，可以买卖两次，也可以不买卖。 +视频讲解：https://www.bilibili.com/video/BV1WG411K7AR +https://programmercarl.com/0123.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAIII.html

+

188.买卖股票的最佳时机IV

+

本题是123.买卖股票的最佳时机III 的进阶版
+视频讲解：https://www.bilibili.com/video/BV16M411U7XJ +https://programmercarl.com/0188.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAIV.html

+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day50/lc123.html b/day50/lc123.html new file mode 100644 index 0000000..b53d395 --- /dev/null +++ b/day50/lc123.html @@ -0,0 +1,246 @@ + + + + + + 123.买卖股票的最佳时机III - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

123.买卖股票的最佳时机III

+

自己写的时候漏掉了dp[0][3]=-prices[0];这个条件

+
class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+//         一天一共就有五个状态，
+
+// 没有操作 （其实我们也可以不设置这个状态）
+// 第一次持有股票
+// 第一次不持有股票
+// 第二次持有股票
+// 第二次不持有股票
+// dp[i][j]中 i表示第i天，j为 [0 - 4] 五个状态，dp[i][j]表示第i天状态j所剩最大现金。
+vector<vector<int>>dp(prices.size(), vector<int>(5,0));
+dp[0][1]=dp[0][3]=-prices[0];
+
+for(int i=1;i<prices.size();i++){
+dp[i][1] = max(dp[i - 1][0] - prices[i], dp[i - 1][1]);
+dp[i][2] = max(dp[i - 1][1] + prices[i], dp[i - 1][2]);
+dp[i][3] = max(dp[i - 1][2] - prices[i], dp[i - 1][3]);
+dp[i][4] = max(dp[i - 1][3] + prices[i], dp[i - 1][4]);
+}
+return max(dp[prices.size()-1][2],dp[prices.size()-1][4]);
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day50/lc188.html b/day50/lc188.html new file mode 100644 index 0000000..77f9f76 --- /dev/null +++ b/day50/lc188.html @@ -0,0 +1,239 @@ + + + + + + 188.买卖股票的最佳时机IV - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

188.买卖股票的最佳时机IV

+

不过度确实好难啊，不过是自己写的，就类比一下上一道题

+
class Solution {
+public:
+    int maxProfit(int k, vector<int>& prices) {
+        vector<vector<int>>dp(prices.size(), vector<int>(2*k+1,0));
+        for(int j=0;j<k;j++)dp[0][2*j+1]=-prices[0];
+        for(int i=1;i<prices.size();i++){
+            for(int j=1;j<=k;j++){
+                dp[i][2*j-1] = max(dp[i - 1][2*j-2] - prices[i], dp[i - 1][2*j-1]);
+                dp[i][2*j]   = max(dp[i - 1][2*j-1] + prices[i], dp[i - 1][2*j]);   // 表示不持有
+            }
+        }
+        int res = -1;
+        for(int j=1;j<=k;j++)res=max(res,dp[prices.size()-1][2*j]);
+        return res;
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day51.html b/day51.html new file mode 100644 index 0000000..9fc9ff8 --- /dev/null +++ b/day51.html @@ -0,0 +1,235 @@ + + + + + + day 51 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

第九章 动态规划part12

+

● 309.最佳买卖股票时机含冷冻期
+● 714.买卖股票的最佳时机含手续费
+●总结

+

309.最佳买卖股票时机含冷冻期

+

本题加了一个冷冻期，状态就多了，有点难度，大家要把各个状态分清，思路才能清晰 +视频讲解：https://www.bilibili.com/video/BV1rP4y1D7ku

+

https://programmercarl.com/0309.%E6%9C%80%E4%BD%B3%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E6%97%B6%E6%9C%BA%E5%90%AB%E5%86%B7%E5%86%BB%E6%9C%9F.html

+

714.买卖股票的最佳时机含手续费

+

相对122.买卖股票的最佳时机II ，本题只需要在计算卖出操作的时候减去手续费就可以了，代码几乎是一样的，可以尝试自己做一做。 +视频讲解：https://www.bilibili.com/video/BV1z44y1Z7UR +https://programmercarl.com/0714.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BA%E5%90%AB%E6%89%8B%E7%BB%AD%E8%B4%B9%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html

+

股票总结

+

股票问题做一个总结吧 +https://programmercarl.com/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92-%E8%82%A1%E7%A5%A8%E9%97%AE%E9%A2%98%E6%80%BB%E7%BB%93%E7%AF%87.html

+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day51/lc309.html b/day51/lc309.html new file mode 100644 index 0000000..fb578fb --- /dev/null +++ b/day51/lc309.html @@ -0,0 +1,240 @@ + + + + + + 309.最佳买卖股票时机含冷冻期 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

309.最佳买卖股票时机含冷冻期

+

我自己定义的状态

+
class Solution {
+public:
+    int maxProfit(vector<int>& v) {
+        // dp[i][j] 表示第i天的最大现金
+        // 0 持有
+        // 1 非持有且不在冷冻期
+        // 2 非持有且冷冻期(刚出售)
+        vector<vector<int>>dp(v.size(),vector<int>(3,0));
+        dp[0][0]=-v[0];
+        for(int i=1;i<v.size();i++){
+            dp[i][0]=max(dp[i-1][0],dp[i-1][1]-v[i]);// 原本就持有 / 解冻后买入
+            dp[i][1]=max(dp[i-1][1],dp[i-1][2]);// 原本就不在冷冻期 / 新的解锁
+            dp[i][2]=dp[i-1][0]+v[i];
+        }
+        return max(dp[v.size()-1][1],dp[v.size()-1][2]);
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day51/lc714.html b/day51/lc714.html new file mode 100644 index 0000000..0f0d9e7 --- /dev/null +++ b/day51/lc714.html @@ -0,0 +1,237 @@ + + + + + + 714.买卖股票的最佳时机含手续费 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

714.买卖股票的最佳时机含手续费

+
class Solution {
+public:
+    int maxProfit(vector<int>& v, int fee) {
+        // dp[i][j]
+        // 0 持有
+        // 1 不持有
+        vector<vector<int>>dp(v.size(),vector<int>(2,0));
+        dp[0][0]=-v[0];
+        for(int i=1;i<v.size();i++){
+            dp[i][0]=max(dp[i-1][0],dp[i-1][1]-v[i]);
+            dp[i][1]=max(dp[i-1][1],dp[i-1][0]+v[i]-fee);
+        }
+        return max(dp[v.size()-1][0],dp[v.size()-1][1]);
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day52.html b/day52.html new file mode 100644 index 0000000..f85828b --- /dev/null +++ b/day52.html @@ -0,0 +1,237 @@ + + + + + + day 52 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

第九章 动态规划part13

+

● 300.最长递增子序列 +● 674. 最长连续递增序列 +● 718. 最长重复子数组

+

详细布置

+

300.最长递增子序列

+

今天开始正式子序列系列，本题是比较简单的，感受感受一下子序列题目的思路。 +视频讲解：https://www.bilibili.com/video/BV1ng411J7xP +https://programmercarl.com/0300.%E6%9C%80%E9%95%BF%E4%B8%8A%E5%8D%87%E5%AD%90%E5%BA%8F%E5%88%97.html

+

674. 最长连续递增序列

+

本题相对于昨天的动态规划：300.最长递增子序列 最大的区别在于“连续”。 先尝试自己做做，感受一下区别
+视频讲解：https://www.bilibili.com/video/BV1bD4y1778v +https://programmercarl.com/0674.%E6%9C%80%E9%95%BF%E8%BF%9E%E7%BB%AD%E9%80%92%E5%A2%9E%E5%BA%8F%E5%88%97.html

+

718. 最长重复子数组

+

稍有难度，要使用二维dp数组了 +视频讲解：https://www.bilibili.com/video/BV178411H7hV +https://programmercarl.com/0718.%E6%9C%80%E9%95%BF%E9%87%8D%E5%A4%8D%E5%AD%90%E6%95%B0%E7%BB%84.html

+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day52/lc300.html b/day52/lc300.html new file mode 100644 index 0000000..e3e2900 --- /dev/null +++ b/day52/lc300.html @@ -0,0 +1,237 @@ + + + + + + 300.最长递增子序列 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

300. 最长递增子序列

+
class Solution {
+public:
+    int lengthOfLIS(vector<int>& v) {
+        // dp[i] 表示以i结尾的LIS的长度
+        int res = 1;
+        vector<int>dp(v.size(),1);
+        for(int i = 1;i<v.size();i++){
+            for(int j = 0;j<i;j++) {
+                if(v[i]>v[j])dp[i]=max(dp[i],dp[j]+1);
+            }
+            res = max(res, dp[i]);
+        }
+        return res;
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day52/lc674.html b/day52/lc674.html new file mode 100644 index 0000000..22356fb --- /dev/null +++ b/day52/lc674.html @@ -0,0 +1,236 @@ + + + + + + 674. 最长连续递增序列 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

674. 最长连续递增序列

+

好像比上一题还简单

+
class Solution {
+public:
+    int findLengthOfLCIS(vector<int>& v) {
+        // dp[i] 表示以i结尾的CLIS的长度
+        int res = 1;
+        vector<int>dp(v.size(),1);
+        for(int i = 1;i<v.size();i++){
+            if(v[i]>v[i-1])dp[i]=max(dp[i],dp[i-1]+1);
+            res = max(res, dp[i]);
+        }
+        return res;
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day52/lc718.html b/day52/lc718.html new file mode 100644 index 0000000..02a1e05 --- /dev/null +++ b/day52/lc718.html @@ -0,0 +1,240 @@ + + + + + + 718. 最长重复子数组 - 代码随想录算法训练营刷题笔记 + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + +
+ +
+ + + + + + + + +
+
+

718. 最长重复子数组

+

抄了随想路

+
class Solution {
+public:
+    int findLength(vector<int>& nums1, vector<int>& nums2) {
+        // dp[i][j] ：以下标i - 1为结尾的A，和以下标j - 1为结尾的B，最长重复子数组长度为dp[i][j]。 （特别注意： “以下标i - 1为结尾的A” 标明一定是 以A[i-1]为结尾的字符串 ）
+        vector<vector<int>> dp (nums1.size() + 1, vector<int>(nums2.size() + 1, 0));
+        int result = 0;
+        for (int i = 1; i <= nums1.size(); i++) {
+            for (int j = 1; j <= nums2.size(); j++) {
+                if (nums1[i - 1] == nums2[j - 1]) {
+                    dp[i][j] = dp[i - 1][j - 1] + 1;
+                }
+                if (dp[i][j] > result) result = dp[i][j];
+            }
+        }
+        return result;
+    }
+};
+
+ +
+ + +
+
+ + + +
+ + + + + + + + + + + + + + + + + + +
+ + diff --git a/day6.html b/day6.html index fa97f43..2ec8b12 100644 --- a/day6.html +++ b/day6.html @@ -83,7 +83,7 @@ diff --git a/day6/lc1.html b/day6/lc1.html index e147407..452609a 100644 --- a/day6/lc1.html +++ b/day6/lc1.html @@ -83,7 +83,7 @@ diff --git a/day6/lc202.html b/day6/lc202.html index 76c17ed..90996b5 100644 --- a/day6/lc202.html +++ b/day6/lc202.html @@ -83,7 +83,7 @@ diff --git a/day6/lc242.html b/day6/lc242.html index 9b1f291..236e257 100644 --- a/day6/lc242.html +++ b/day6/lc242.html @@ -83,7 +83,7 @@ diff --git a/day6/lc349.html b/day6/lc349.html index 51adc58..194086d 100644 --- a/day6/lc349.html +++ b/day6/lc349.html @@ -83,7 +83,7 @@ diff --git a/day7.html b/day7.html index 7611041..1d2751b 100644 --- a/day7.html +++ b/day7.html @@ -83,7 +83,7 @@ diff --git a/day7/lc15.html b/day7/lc15.html index 945fe68..7ce8eec 100644 --- a/day7/lc15.html +++ b/day7/lc15.html @@ -83,7 +83,7 @@ diff --git a/day7/lc18.html b/day7/lc18.html index bacad83..0de7916 100644 --- a/day7/lc18.html +++ b/day7/lc18.html @@ -83,7 +83,7 @@ diff --git a/day7/lc383.html b/day7/lc383.html index 2e04584..84f0a99 100644 --- a/day7/lc383.html +++ b/day7/lc383.html @@ -83,7 +83,7 @@ diff --git a/day7/lc454.html b/day7/lc454.html index 430aab8..2cbae91 100644 --- a/day7/lc454.html +++ b/day7/lc454.html @@ -83,7 +83,7 @@ diff --git a/day8.html b/day8.html index 0cfbf3b..1d0f770 100644 --- a/day8.html +++ b/day8.html @@ -83,7 +83,7 @@ diff --git a/day8/lc151.html b/day8/lc151.html index 0a2b7eb..6a654eb 100644 --- a/day8/lc151.html +++ b/day8/lc151.html @@ -83,7 +83,7 @@ diff --git a/day8/lc344.html b/day8/lc344.html index 1640bc4..57f1490 100644 --- a/day8/lc344.html +++ b/day8/lc344.html @@ -83,7 +83,7 @@ diff --git a/day8/lc541.html b/day8/lc541.html index 96d230a..484d0e1 100644 --- a/day8/lc541.html +++ b/day8/lc541.html @@ -83,7 +83,7 @@ diff --git a/day8/offer5.html b/day8/offer5.html index 12ed7ff..dd321ac 100644 --- a/day8/offer5.html +++ b/day8/offer5.html @@ -83,7 +83,7 @@ diff --git a/day8/offer58.html b/day8/offer58.html index f3002a6..b902fa3 100644 --- a/day8/offer58.html +++ b/day8/offer58.html @@ -83,7 +83,7 @@ diff --git a/day9.html b/day9.html index 746304d..d08e17f 100644 --- a/day9.html +++ b/day9.html @@ -83,7 +83,7 @@ diff --git a/day9/lc28.html b/day9/lc28.html index 93110f3..2ea18df 100644 --- a/day9/lc28.html +++ b/day9/lc28.html @@ -83,7 +83,7 @@ diff --git a/day9/lc459.html b/day9/lc459.html index 1e45920..0398a46 100644 --- a/day9/lc459.html +++ b/day9/lc459.html @@ -83,7 +83,7 @@ diff --git a/index.html b/index.html index 1d0ccce..444e389 100644 --- a/index.html +++ b/index.html @@ -83,7 +83,7 @@ diff --git a/print.html b/print.html index abc5d04..deddf2f 100644 --- a/print.html +++ b/print.html @@ -84,7 +84,7 @@ @@ -4491,6 +4491,364 @@

背包问题总结篇！

https://programmercarl.com/%E8%83%8C%E5%8C%85%E6%80%BB%E7%BB%93%E7%AF%87.html

+

139. 单词拆分

+

不会

+
class Solution {
+public:
+    bool wordBreak(string s, vector<string>& wordDict) {
+        // dp[i] -> 长度为i的字符串是否可以拼出来
+        // dp[j] = true if dp[j-i] == true and [i:j] in wordDict
+        unordered_set<string> wordSet(wordDict.begin(),wordDict.end());
+        vector<bool>dp(s.size()+1,false);
+        dp[0]=true;
+        for (int i = 1 ; i<=s.size();i++) {
+            for (int j = 0; j < i; j ++ ) {
+                string word = s.substr(j,i-j);
+                if (dp[j] && wordSet.find(word)!= wordSet.end())
+                    dp[i] = true;
+            }
+        }
+        return dp[s.size()];
+    }
+};
+
+

第九章 动态规划part09

+

● 198.打家劫舍 +● 213.打家劫舍II
+● 337.打家劫舍III

+

详细布置

+

今天就是打家劫舍的一天，这个系列不算难，大家可以一口气拿下。

+

198.打家劫舍

+

视频讲解：https://www.bilibili.com/video/BV1Te411N7SX +https://programmercarl.com/0198.%E6%89%93%E5%AE%B6%E5%8A%AB%E8%88%8D.html

+

213.打家劫舍II

+

视频讲解：https://www.bilibili.com/video/BV1oM411B7xq +https://programmercarl.com/0213.%E6%89%93%E5%AE%B6%E5%8A%AB%E8%88%8DII.html

+

337.打家劫舍III

+

视频讲解：https://www.bilibili.com/video/BV1H24y1Q7sY +https://programmercarl.com/0337.%E6%89%93%E5%AE%B6%E5%8A%AB%E8%88%8DIII.html

+

198. 打家劫舍

+

这个是抄随想录的

+
class Solution {
+public:
+    int rob(vector<int>& nums) {
+        if (nums.size() == 0) return 0;
+        if (nums.size() == 1) return nums[0];
+        // dp[i]：考虑下标i（包括i）以内的房屋，最多可以偷窃的金额为dp[i]
+        // dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        vector<int> dp(nums.size());
+        dp[0] = nums[0];
+        dp[1] = max(nums[0], nums[1]);
+        for (int i = 2; i < nums.size(); i++) {
+            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+        return dp[nums.size() - 1];
+    }
+};
+
+

213. 打家劫舍 II

+

这个是自己写的

+
class Solution {
+public:
+    int rob(vector<int>& nums) {
+        if (nums.size() == 0) return 0;
+        if (nums.size() == 1) return nums[0];
+        // dp[i]：考虑下标i（包括i）以内的房屋，最多可以偷窃的金额为dp[i]
+        // dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        vector<int> dp(nums.size());
+        // skip index 0 first
+        int res = 0;
+        dp[0] = 0;
+        dp[1] = nums[1];
+        for (int i = 2; i < nums.size(); i++) {
+            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+        res = dp[nums.size()-1];
+        dp[0] = nums[0];
+        dp[1] = max(nums[0], nums[1]);
+        
+        for (int i = 2; i < nums.size() - 1; i++) {
+            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+
+        res = max(res,dp[nums.size()-2]);
+        return res;
+    }
+};
+
+

337.打家劫舍III

+

自己做的

+
/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+// <r1,r2> r1 如果偷当前节点，得到的最大值 ； r2 不偷当前节点的最大值
+    pair<int,int> f(TreeNode*root) {
+        if(!root){return make_pair(0,0);}
+        auto leftr = f(root->left);
+        auto rightr = f(root->right);
+        int takeleft = leftr.first;
+        int notakeleft = leftr.second;
+        int takeright = rightr.first;
+        int notakeright = rightr.second;
+        int r1 = notakeleft + notakeright + root->val;
+        int r2 = max(takeleft,notakeleft) + max(takeright,notakeright);
+        return make_pair(r1, r2);
+    }
+    int rob(TreeNode* root) {
+        auto res = f(root);
+        return max(res.first, res.second);
+    }
+};
+
+

第九章 动态规划part10

+

● 121. 买卖股票的最佳时机 +● 122.买卖股票的最佳时机II

+

详细布置

+

股票问题是一个动态规划的系列问题，今日安排的题目不多，大家可以慢慢消化。

+

121. 买卖股票的最佳时机

+

视频讲解：https://www.bilibili.com/video/BV1Xe4y1u77q +https://programmercarl.com/0121.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BA.html

+

122.买卖股票的最佳时机II

+

视频讲解：https://www.bilibili.com/video/BV1D24y1Q7Ls +https://programmercarl.com/0122.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAII%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html

+

121. 买卖股票的最佳时机

+

一开始这样子没有用动规

+
class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int res = 0;
+        int cmin = INT_MAX;
+        for (int i = 0; i < prices.size(); i ++ ) {
+            cmin = min(cmin, prices[i]);
+            if (prices[i] > cmin) {
+                res = max(res, prices[i] - cmin);
+            }
+        }
+        return res;
+    }
+};
+
+

122. 买卖股票的最佳时机 II

+

抄了随想录

+
class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int len = prices.size();
+        vector<vector<int>> dp(len, vector<int>(2, 0));
+        // dp[i][0] 表示第i天持有股票所得现金。
+        // dp[i][1] 表示第i天不持有股票所得最多现金
+
+// 如果第i天持有股票即dp[i][0]， 那么可以由两个状态推出来
+
+// 第i-1天就持有股票，那么就保持现状，所得现金就是昨天持有股票的所得现金 即：dp[i - 1][0]
+// 第i天买入股票，所得现金就是昨天不持有股票的所得现金减去 今天的股票价格 即：dp[i - 1][1] - prices[i]
+
+
+// 再来看看如果第i天不持有股票即dp[i][1]的情况， 依然可以由两个状态推出来
+
+// 第i-1天就不持有股票，那么就保持现状，所得现金就是昨天不持有股票的所得现金 即：dp[i - 1][1]
+// 第i天卖出股票，所得现金就是按照今天股票价格卖出后所得现金即：prices[i] + dp[i - 1][0]
+        dp[0][0] -= prices[0];
+        dp[0][1] = 0;
+        for (int i = 1; i < len; i++) {
+            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]); // 注意这里是和121. 买卖股票的最佳时机唯一不同的地方。
+            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i]);
+        }
+        return dp[len - 1][1];
+
+
+    }
+};
+
+

第九章 动态规划part11

+

● 123.买卖股票的最佳时机III
+● 188.买卖股票的最佳时机IV

+

详细布置

+

123.买卖股票的最佳时机III

+

这道题一下子就难度上来了，关键在于至多买卖两次，这意味着可以买卖一次，可以买卖两次，也可以不买卖。 +视频讲解：https://www.bilibili.com/video/BV1WG411K7AR +https://programmercarl.com/0123.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAIII.html

+

188.买卖股票的最佳时机IV

+

本题是123.买卖股票的最佳时机III 的进阶版
+视频讲解：https://www.bilibili.com/video/BV16M411U7XJ +https://programmercarl.com/0188.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAIV.html

+

123.买卖股票的最佳时机III

+

自己写的时候漏掉了dp[0][3]=-prices[0];这个条件

+
class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+//         一天一共就有五个状态，
+
+// 没有操作 （其实我们也可以不设置这个状态）
+// 第一次持有股票
+// 第一次不持有股票
+// 第二次持有股票
+// 第二次不持有股票
+// dp[i][j]中 i表示第i天，j为 [0 - 4] 五个状态，dp[i][j]表示第i天状态j所剩最大现金。
+vector<vector<int>>dp(prices.size(), vector<int>(5,0));
+dp[0][1]=dp[0][3]=-prices[0];
+
+for(int i=1;i<prices.size();i++){
+dp[i][1] = max(dp[i - 1][0] - prices[i], dp[i - 1][1]);
+dp[i][2] = max(dp[i - 1][1] + prices[i], dp[i - 1][2]);
+dp[i][3] = max(dp[i - 1][2] - prices[i], dp[i - 1][3]);
+dp[i][4] = max(dp[i - 1][3] + prices[i], dp[i - 1][4]);
+}
+return max(dp[prices.size()-1][2],dp[prices.size()-1][4]);
+    }
+};
+
+

188.买卖股票的最佳时机IV

+

不过度确实好难啊，不过是自己写的，就类比一下上一道题

+
class Solution {
+public:
+    int maxProfit(int k, vector<int>& prices) {
+        vector<vector<int>>dp(prices.size(), vector<int>(2*k+1,0));
+        for(int j=0;j<k;j++)dp[0][2*j+1]=-prices[0];
+        for(int i=1;i<prices.size();i++){
+            for(int j=1;j<=k;j++){
+                dp[i][2*j-1] = max(dp[i - 1][2*j-2] - prices[i], dp[i - 1][2*j-1]);
+                dp[i][2*j]   = max(dp[i - 1][2*j-1] + prices[i], dp[i - 1][2*j]);   // 表示不持有
+            }
+        }
+        int res = -1;
+        for(int j=1;j<=k;j++)res=max(res,dp[prices.size()-1][2*j]);
+        return res;
+    }
+};
+
+

第九章 动态规划part12

+

● 309.最佳买卖股票时机含冷冻期
+● 714.买卖股票的最佳时机含手续费
+●总结

+

309.最佳买卖股票时机含冷冻期

+

本题加了一个冷冻期，状态就多了，有点难度，大家要把各个状态分清，思路才能清晰 +视频讲解：https://www.bilibili.com/video/BV1rP4y1D7ku

+

https://programmercarl.com/0309.%E6%9C%80%E4%BD%B3%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E6%97%B6%E6%9C%BA%E5%90%AB%E5%86%B7%E5%86%BB%E6%9C%9F.html

+

714.买卖股票的最佳时机含手续费

+

相对122.买卖股票的最佳时机II ，本题只需要在计算卖出操作的时候减去手续费就可以了，代码几乎是一样的，可以尝试自己做一做。 +视频讲解：https://www.bilibili.com/video/BV1z44y1Z7UR +https://programmercarl.com/0714.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BA%E5%90%AB%E6%89%8B%E7%BB%AD%E8%B4%B9%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html

+

股票总结

+

股票问题做一个总结吧 +https://programmercarl.com/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92-%E8%82%A1%E7%A5%A8%E9%97%AE%E9%A2%98%E6%80%BB%E7%BB%93%E7%AF%87.html

+

309.最佳买卖股票时机含冷冻期

+

我自己定义的状态

+
class Solution {
+public:
+    int maxProfit(vector<int>& v) {
+        // dp[i][j] 表示第i天的最大现金
+        // 0 持有
+        // 1 非持有且不在冷冻期
+        // 2 非持有且冷冻期(刚出售)
+        vector<vector<int>>dp(v.size(),vector<int>(3,0));
+        dp[0][0]=-v[0];
+        for(int i=1;i<v.size();i++){
+            dp[i][0]=max(dp[i-1][0],dp[i-1][1]-v[i]);// 原本就持有 / 解冻后买入
+            dp[i][1]=max(dp[i-1][1],dp[i-1][2]);// 原本就不在冷冻期 / 新的解锁
+            dp[i][2]=dp[i-1][0]+v[i];
+        }
+        return max(dp[v.size()-1][1],dp[v.size()-1][2]);
+    }
+};
+
+

714.买卖股票的最佳时机含手续费

+
class Solution {
+public:
+    int maxProfit(vector<int>& v, int fee) {
+        // dp[i][j]
+        // 0 持有
+        // 1 不持有
+        vector<vector<int>>dp(v.size(),vector<int>(2,0));
+        dp[0][0]=-v[0];
+        for(int i=1;i<v.size();i++){
+            dp[i][0]=max(dp[i-1][0],dp[i-1][1]-v[i]);
+            dp[i][1]=max(dp[i-1][1],dp[i-1][0]+v[i]-fee);
+        }
+        return max(dp[v.size()-1][0],dp[v.size()-1][1]);
+    }
+};
+
+

第九章 动态规划part13

+

● 300.最长递增子序列 +● 674. 最长连续递增序列 +● 718. 最长重复子数组

+

详细布置

+

300.最长递增子序列

+

今天开始正式子序列系列，本题是比较简单的，感受感受一下子序列题目的思路。 +视频讲解：https://www.bilibili.com/video/BV1ng411J7xP +https://programmercarl.com/0300.%E6%9C%80%E9%95%BF%E4%B8%8A%E5%8D%87%E5%AD%90%E5%BA%8F%E5%88%97.html

+

674. 最长连续递增序列

+

本题相对于昨天的动态规划：300.最长递增子序列 最大的区别在于“连续”。 先尝试自己做做，感受一下区别
+视频讲解：https://www.bilibili.com/video/BV1bD4y1778v +https://programmercarl.com/0674.%E6%9C%80%E9%95%BF%E8%BF%9E%E7%BB%AD%E9%80%92%E5%A2%9E%E5%BA%8F%E5%88%97.html

+

718. 最长重复子数组

+

稍有难度，要使用二维dp数组了 +视频讲解：https://www.bilibili.com/video/BV178411H7hV +https://programmercarl.com/0718.%E6%9C%80%E9%95%BF%E9%87%8D%E5%A4%8D%E5%AD%90%E6%95%B0%E7%BB%84.html

+

300. 最长递增子序列

+
class Solution {
+public:
+    int lengthOfLIS(vector<int>& v) {
+        // dp[i] 表示以i结尾的LIS的长度
+        int res = 1;
+        vector<int>dp(v.size(),1);
+        for(int i = 1;i<v.size();i++){
+            for(int j = 0;j<i;j++) {
+                if(v[i]>v[j])dp[i]=max(dp[i],dp[j]+1);
+            }
+            res = max(res, dp[i]);
+        }
+        return res;
+    }
+};
+
+

674. 最长连续递增序列

+

好像比上一题还简单

+
class Solution {
+public:
+    int findLengthOfLCIS(vector<int>& v) {
+        // dp[i] 表示以i结尾的CLIS的长度
+        int res = 1;
+        vector<int>dp(v.size(),1);
+        for(int i = 1;i<v.size();i++){
+            if(v[i]>v[i-1])dp[i]=max(dp[i],dp[i-1]+1);
+            res = max(res, dp[i]);
+        }
+        return res;
+    }
+};
+
+

718. 最长重复子数组

+

抄了随想路

+
class Solution {
+public:
+    int findLength(vector<int>& nums1, vector<int>& nums2) {
+        // dp[i][j] ：以下标i - 1为结尾的A，和以下标j - 1为结尾的B，最长重复子数组长度为dp[i][j]。 （特别注意： “以下标i - 1为结尾的A” 标明一定是 以A[i-1]为结尾的字符串 ）
+        vector<vector<int>> dp (nums1.size() + 1, vector<int>(nums2.size() + 1, 0));
+        int result = 0;
+        for (int i = 1; i <= nums1.size(); i++) {
+            for (int j = 1; j <= nums2.size(); j++) {
+                if (nums1[i - 1] == nums2[j - 1]) {
+                    dp[i][j] = dp[i - 1][j - 1] + 1;
+                }
+                if (dp[i][j] > result) result = dp[i][j];
+            }
+        }
+        return result;
+    }
+};
+
diff --git a/remains.html b/remains.html index 02c8185..bb763d7 100644 --- a/remains.html +++ b/remains.html @@ -83,7 +83,7 @@ @@ -171,7 +171,7 @@

代码随想录算法训练营刷题笔记