From 2af9c98e5c07e4eaca340effee25be02c161f723 Mon Sep 17 00:00:00 2001 From: "zhipengxu (aka. Ainevsia)" Date: Fri, 25 Aug 2023 21:37:10 +0800 Subject: [PATCH] ;1 --- notes/src/SUMMARY.md | 11 +++++++++++ notes/src/day43.md | 23 +++++++++++++++++++++++ notes/src/day43/lc1049.md | 21 +++++++++++++++++++++ notes/src/day43/lc474.md | 27 +++++++++++++++++++++++++++ notes/src/day43/lc494.md | 27 +++++++++++++++++++++++++++ notes/src/day44.md | 22 ++++++++++++++++++++++ notes/src/day44/lc377.md | 21 +++++++++++++++++++++ notes/src/day44/lc518.md | 21 +++++++++++++++++++++ notes/src/day45.md | 27 +++++++++++++++++++++++++++ notes/src/day45/lc279.md | 21 +++++++++++++++++++++ notes/src/day45/lc322.md | 24 ++++++++++++++++++++++++ notes/src/day45/lc70.md | 20 ++++++++++++++++++++ 12 files changed, 265 insertions(+) create mode 100644 notes/src/day43.md create mode 100644 notes/src/day43/lc1049.md create mode 100644 notes/src/day43/lc474.md create mode 100644 notes/src/day43/lc494.md create mode 100644 notes/src/day44.md create mode 100644 notes/src/day44/lc377.md create mode 100644 notes/src/day44/lc518.md create mode 100644 notes/src/day45.md create mode 100644 notes/src/day45/lc279.md create mode 100644 notes/src/day45/lc322.md create mode 100644 notes/src/day45/lc70.md diff --git a/notes/src/SUMMARY.md b/notes/src/SUMMARY.md index d441aa1..b111590 100644 --- a/notes/src/SUMMARY.md +++ b/notes/src/SUMMARY.md @@ -139,4 +139,15 @@ - [96.不同的二叉搜索树](./day41/lc96.md) - [day 42](./day42.md) - [416. 分割等和子集](./day42/lc416.md) +- [day 43](./day43.md) + - [1049. 最后一块石头的重量 II](./day43/lc1049.md) + - [494. 目标和](./day43/lc494.md) + - [474. 一和零](./day43/lc474.md) +- [day 44](./day44.md) + - [518. 零钱兑换 II](./day44/lc518.md) + - [377. 组合总和 Ⅳ](./day44/lc377.md) +- [day 45](./day45.md) + - [70. 爬楼梯 （进阶）](./day45/lc70.md) + - [322. 零钱兑换](./day45/lc322.md) + - [279.完全平方数](./day45/lc279.md) - [remains](./remains.md) \ No newline at end of file diff --git a/notes/src/day43.md b/notes/src/day43.md new file mode 100644 index 0000000..9780fe6 --- /dev/null +++ b/notes/src/day43.md @@ -0,0 +1,23 @@ +# 第九章 动态规划 part05 +● 1049. 最后一块石头的重量 II +● 494. 目标和 +● 474.一和零 + + 详细布置 +## 1049. 最后一块石头的重量 II + +本题就和 昨天的 416. 分割等和子集 很像了，可以尝试先自己思考做一做。 +视频讲解：https://www.bilibili.com/video/BV14M411C7oV +https://programmercarl.com/1049.%E6%9C%80%E5%90%8E%E4%B8%80%E5%9D%97%E7%9F%B3%E5%A4%B4%E7%9A%84%E9%87%8D%E9%87%8FII.html +## 494. 目标和 +大家重点理解 递推公式：dp[j] += dp[j - nums[i]]，这个公式后面的提问 我们还会用到。 +视频讲解：https://www.bilibili.com/video/BV1o8411j73x +https://programmercarl.com/0494.%E7%9B%AE%E6%A0%87%E5%92%8C.html + +## 474.一和零 +通过这道题目，大家先粗略了解， 01背包，完全背包，多重背包的区别，不过不用细扣，因为后面 对于 完全背包，多重背包 还有单独讲解。 +视频讲解：https://www.bilibili.com/video/BV1rW4y1x7ZQ +https://programmercarl.com/0474.%E4%B8%80%E5%92%8C%E9%9B%B6.html + + + diff --git a/notes/src/day43/lc1049.md b/notes/src/day43/lc1049.md new file mode 100644 index 0000000..c5a4690 --- /dev/null +++ b/notes/src/day43/lc1049.md @@ -0,0 +1,21 @@ +# 1049. 最后一块石头的重量 II + +```cpp +class Solution { +public: + int lastStoneWeightII(vector& nums) { + int sum = 0; + for (int i = 0; i < nums.size(); i ++ ) { + sum += nums[i]; + } + int target = sum >> 1; + vector dp (1+target,0); + for (int i = 0; i < nums.size(); i ++ ) { + for (int j = target; j >= nums[i]; j -- ) { + dp[j] = max(dp[j],dp[j-nums[i]]+nums[i]); + } + } + return sum - dp[target]*2; + } +}; +``` \ No newline at end of file diff --git a/notes/src/day43/lc474.md b/notes/src/day43/lc474.md new file mode 100644 index 0000000..094fd2b --- /dev/null +++ b/notes/src/day43/lc474.md @@ -0,0 +1,27 @@ +# 474.一和零 + +```cpp +class Solution { +public: + int findMaxForm(vector& strs, int m, int n) { + // dp[i][j] 表示子集中最多i个0和j个1的最大子集长度 + vector>dp(1+m,vector(1+n,0)); + // x = count(s.begin(),s.end(),'0'); + // y = s.size() - x; + // dp[i][j] = max(dp[i][j],dp[i-x][j-y]+1) + for (int u = 0; u < strs.size() ; u ++ ) { + string s = strs[u]; + int x = count(s.begin(),s.end(),'0'); + int y = s.size() - x; + for ( int i = m ; i >= x; i -- ) { + for ( int j = n ; j >= y; j -- ) { + dp[i][j] = max(dp[i][j],dp[i-x][j-y]+1); + } + } + } + return dp[m][n]; + } +}; +``` + +这个题很舒服，自己一下写过的 \ No newline at end of file diff --git a/notes/src/day43/lc494.md b/notes/src/day43/lc494.md new file mode 100644 index 0000000..987ea65 --- /dev/null +++ b/notes/src/day43/lc494.md @@ -0,0 +1,27 @@ +# 494.目标和 + +一开始不会做，一种心的dp推导公式 + +```cpp +class Solution { +public: + int findTargetSumWays(vector& nums, int target) { + int sum = 0; + for (int i = 0; i < nums.size(); i ++ ) { + sum += nums[i]; + } + int left = (target + sum); + if (left & 1) return 0; + left = left >> 1; + vectordp(left+1); + dp[0] = 1; + // dp[j] 表示：填满j（包括j）这么大容积的包，有dp[j]种方法 + for ( int i = 0; i < nums.size() ; i ++ ) { + for (int j = dp.size() - 1; j >= nums[i] ; j -- ) { + dp[j] += dp[j-nums[i]]; + } + } + return dp[left]; + } +}; +``` \ No newline at end of file diff --git a/notes/src/day44.md b/notes/src/day44.md new file mode 100644 index 0000000..e2f855b --- /dev/null +++ b/notes/src/day44.md @@ -0,0 +1,22 @@ +# 第九章 动态规划part06 +● 完全背包 +● 518. 零钱兑换 II +● 377. 组合总和 Ⅳ + + 详细布置 + +力扣上没有纯粹的完全背包的题目，所以大家看本篇了解一下 完全背包的理论 + +后面的两道题目，都是完全背包的应用，做做感受一下 + +## 完全背包 +视频讲解：https://www.bilibili.com/video/BV1uK411o7c9 +https://programmercarl.com/%E8%83%8C%E5%8C%85%E9%97%AE%E9%A2%98%E7%90%86%E8%AE%BA%E5%9F%BA%E7%A1%80%E5%AE%8C%E5%85%A8%E8%83%8C%E5%8C%85.html + +## 518. 零钱兑换 II +视频讲解：https://www.bilibili.com/video/BV1KM411k75j +https://programmercarl.com/0518.%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2II.html + +## 377. 组合总和 Ⅳ +视频讲解：https://www.bilibili.com/video/BV1V14y1n7B6 +https://programmercarl.com/0377.%E7%BB%84%E5%90%88%E6%80%BB%E5%92%8C%E2%85%A3.html diff --git a/notes/src/day44/lc377.md b/notes/src/day44/lc377.md new file mode 100644 index 0000000..b259281 --- /dev/null +++ b/notes/src/day44/lc377.md @@ -0,0 +1,21 @@ +# 377. 组合总和 Ⅳ + +不会做，抄了随想录 + +```cpp +class Solution { +public: + int combinationSum4(vector& nums, int target) { + // dp[i] 表示总和为i的元素组合的个数 + vectordp(1+target); + dp[0]=1; + for(int j=0;j<=target;j++){ + for(int i=0;i=nums[i]) + dp[j]+=dp[j-nums[i]]; + } + } + return dp[target]; + } +}; +``` \ No newline at end of file diff --git a/notes/src/day44/lc518.md b/notes/src/day44/lc518.md new file mode 100644 index 0000000..e5855a6 --- /dev/null +++ b/notes/src/day44/lc518.md @@ -0,0 +1,21 @@ +# 518. 零钱兑换 II + +第一道完全背包的题目，抄了随想录 + +```cpp +class Solution { +public: + int change(int amount, vector& coins) { + // dp[j]：凑成总金额j的货币组合数为dp[j] + vectordp(amount+1); + dp[0]=1; + // dp[j] += dp[j - coins[i]]; + for(int i = 0; i < coins.size(); i ++ ) { + for (int j = coins[i]; j <= amount; j ++ ) { // 完全背包，需要重复计算之前的组合数 + dp[j] += dp[j-coins[i]]; + } + } + return dp[amount]; + } +}; +``` \ No newline at end of file diff --git a/notes/src/day45.md b/notes/src/day45.md new file mode 100644 index 0000000..f99660e --- /dev/null +++ b/notes/src/day45.md @@ -0,0 +1,27 @@ +# 第九章 动态规划part07 + +● 70. 爬楼梯 （进阶） +● 322. 零钱兑换 +● 279.完全平方数 + + 详细布置 + +## 70. 爬楼梯 （进阶） + +这道题目 爬楼梯之前我们做过，这次再用完全背包的思路来分析一遍 + +https://programmercarl.com/0070.%E7%88%AC%E6%A5%BC%E6%A2%AF%E5%AE%8C%E5%85%A8%E8%83%8C%E5%8C%85%E7%89%88%E6%9C%AC.html + +## 322. 零钱兑换 + +如果求组合数就是外层for循环遍历物品，内层for遍历背包。 +如果求排列数就是外层for遍历背包，内层for循环遍历物品。 + +这句话结合本题 大家要好好理解。 +视频讲解：https://www.bilibili.com/video/BV14K411R7yv +https://programmercarl.com/0322.%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2.html + +## 279.完全平方数 +本题 和 322. 零钱兑换 基本是一样的，大家先自己尝试做一做 +视频讲解：https://www.bilibili.com/video/BV12P411T7Br +https://programmercarl.com/0279.%E5%AE%8C%E5%85%A8%E5%B9%B3%E6%96%B9%E6%95%B0.html diff --git a/notes/src/day45/lc279.md b/notes/src/day45/lc279.md new file mode 100644 index 0000000..8af84b8 --- /dev/null +++ b/notes/src/day45/lc279.md @@ -0,0 +1,21 @@ +# 279.完全平方数 + + +喜喜这个自己做的 +```cpp +class Solution { +public: + int numSquares(int n) { + // dp[i] 和为 i 的完全平方数的最少数量 + // dp[j] = min(dp[j],dp[j-i*i]+1) + vectordp(1+n,INT_MAX); + dp[0]=0; + for(int i = 1; i * i <= n; i ++ ) { + for(int j = i*i; j <= n; j ++ ) { + dp[j] = min(dp[j],dp[j-i*i]+1); + } + } + return dp[n]; + } +}; +``` \ No newline at end of file diff --git a/notes/src/day45/lc322.md b/notes/src/day45/lc322.md new file mode 100644 index 0000000..abd8894 --- /dev/null +++ b/notes/src/day45/lc322.md @@ -0,0 +1,24 @@ +# 322. 零钱兑换 + +抄了随想录 + +```cpp +class Solution { +public: + int coinChange(vector& coins, int amount) { + // if (amount == 0) return 0; + // dp[i] -> 可以凑成i所需的 最少的硬币个数 + // dp[i] = min(dp[i],dp[i-coins[j]]+1) + vectordp(amount+1,INT_MAX); + dp[0]=0; + for(int i=0;i 爬到有i个台阶的楼顶，有dp[i]种方法 + vector dp(1+n); + dp[0]=1; + for(int i=0;i<=n;i++) { + for(int j=1;j<=m;j++) { + if(i-j>=0) + dp[i]+=dp[i-j]; + } + } + return dp[n]; + } +}; +``` \ No newline at end of file