一刷474

README.adoc

@@ -3343,14 +3343,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/0473-matchsticks-to-square.adoc[题解]
 //|Medium
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/ones-and-zeroes/[474. Ones and Zeroes^]
-//|{source_base_url}/_0474_OnesAndZeroes.java[Java]
-//|{doc_base_url}/0474-ones-and-zeroes.adoc[题解]
-//|Medium
-//|
-//
+
+|{counter:codes}
+|{leetcode_base_url}/ones-and-zeroes/[474. Ones and Zeroes^]
+|{source_base_url}/_0474_OnesAndZeroes.java[Java]
+|{doc_base_url}/0474-ones-and-zeroes.adoc[题解]
+|Medium
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/heaters/[475. Heaters^]
 //|{source_base_url}/_0475_Heaters.java[Java]

docs/0474-ones-and-zeroes.adoc

@@ -1,52 +1,77 @@
 [#0474-ones-and-zeroes]
-= 474. Ones and Zeroes
+= 474. 一和零
 
-{leetcode}/problems/ones-and-zeroes/[LeetCode - Ones and Zeroes^]
+https://leetcode.cn/problems/ones-and-zeroes/[LeetCode - 474. 一和零 ^]
 
-In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.
+给你一个二进制字符串数组 `+strs+` 和两个整数 `+m+` 和 `+n+` 。
 
-For now, suppose you are a dominator of *m* `0s` and *n* `1s` respectively. On the other hand, there is an array with strings consisting of only `0s` and `1s`.
+请你找出并返回 `+strs+` 的最大子集的长度，该子集中 *最多* 有 `+m+` 个 `+0+` 和 `+n+` 个 `+1+` 。
 
-Now your task is to find the maximum number of strings that you can form with given *m* `0s` and *n* `1s`. Each `0` and `1` can be used at most *once*.
+如果 `+x+` 的所有元素也是 `+y+` 的元素，集合 `+x+` 是集合 `+y+` 的 *子集* 。
 
-*Note:*
+*示例 1：*
 
+....
+输入：strs = ["10", "0001", "111001", "1", "0"], m = 5, n = 3
+输出：4
+解释：最多有 5 个 0 和 3 个 1 的最大子集是 {"10","0001","1","0"} ，因此答案是 4 。
+其他满足题意但较小的子集包括 {"0001","1"} 和 {"10","1","0"} 。{"111001"} 不满足题意，因为它含 4 个 1 ，大于 n 的值 3 。
+....
 
-. The given numbers of `0s` and `1s` will both not exceed `100`
-. The size of given string array won't exceed `600`.
+*示例 2：*
 
+....
+输入：strs = ["10", "0", "1"], m = 1, n = 1
+输出：2
+解释：最大的子集是 {"0", "1"} ，所以答案是 2 。
+....
 
-
 
-*Example 1:*
+*提示：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
-*Output:* 4
-
-*Explanation:* This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”
-----
+* `+1 <= strs.length <= 600+`
+* `+1 <= strs[i].length <= 100+`
+* `+strs[i]+` 仅由 `+'0'+` 和 `+'1'+` 组成
+* `+1 <= m, n <= 100+`
 
-
 
-*Example 2:*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* Array = {"10", "0", "1"}, m = 1, n = 1
-*Output:* 2
+== 思路分析
 
-*Explanation:* You could form "10", but then you'd have nothing left. Better form "0" and "1".
-----
+思路：把总共的 0 和 1 的个数视为背包的容量，每一个字符串视为装进背包的物品。这道题就可以使用 0-1 背包问题的思路完成，这里的目标值是能放进背包的字符串的数量。
 
- 
+物品一个一个尝试，容量一点一点尝试，每个物品分类讨论的标准是：选与不选。
 
+dp[i−1][j][k]  // 不选择当前考虑的字符串，至少是这个数值
 
+dp[i−1][j−当前字符串使用0的个数][k−当前字符串使用1的个数] + 1  // 选择当前考虑的字符串
 
 [[src-0474]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0474_OnesAndZeroes.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0474_OnesAndZeroes_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
+
+. https://leetcode.cn/problems/ones-and-zeroes/solutions/814806/yi-he-ling-by-leetcode-solution-u2z2/[474. 一和零 - 官方题解^]
+
+
 

docs/index.adoc

@@ -1022,7 +1022,7 @@ include::0462-minimum-moves-to-equal-array-elements-ii.adoc[leveloffset=+1]
 
 // include::0473-matchsticks-to-square.adoc[leveloffset=+1]
 
-// include::0474-ones-and-zeroes.adoc[leveloffset=+1]
+include::0474-ones-and-zeroes.adoc[leveloffset=+1]
 
 // include::0475-heaters.adoc[leveloffset=+1]
 

logbook/202406.adoc

@@ -929,6 +929,12 @@
 |{doc_base_url}/0031-next-permutation.adoc[题解]
 |❌ 一脸懵逼，可以看成寻找下一个更大的数。这样更容易理解。
 
+|{counter:codes}
+|{leetcode_base_url}/ones-and-zeroes/[474. Ones and Zeroes^]
+|{doc_base_url}/0474-ones-and-zeroes.adoc[题解]
+|❌ 动态规划，0-1 背包问题。
+
+
 |===
 
 截止目前，本轮练习一共完成 {codes} 道题。

src/main/java/com/diguage/algo/leetcode/_0474_OnesAndZeroes.java

@@ -0,0 +1,39 @@
+package com.diguage.algo.leetcode;
+
+public class _0474_OnesAndZeroes {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-10-22 19:47:11
+   */
+  public int findMaxForm(String[] strs, int m, int n) {
+    int length = strs.length;
+    int[][] bits = new int[2][length];
+    for (int i = 0; i < length; i++) {
+      for (int j = 0; j < strs[i].length(); j++) {
+        char c = strs[i].charAt(j);
+        bits[c - '0'][i]++;
+      }
+    }
+    // 确定 dp 数组（dp table）以及下标的含义
+    // dp 数组如何初始化
+    int[][][] dp = new int[length + 1][m + 1][n + 1];
+    // 确定遍历顺序
+    for (int i = 1; i <= length; i++) {
+      int zero = bits[0][i - 1];
+      int one = bits[1][i - 1];
+      for (int j = 0; j <= m; j++) {
+        for (int k = 0; k <= n; k++) {
+          // 确定递推公式
+          dp[i][j][k] = dp[i - 1][j][k];
+          if (j >= zero && k >= one) {
+            dp[i][j][k] = Math.max(dp[i][j][k], dp[i - 1][j - zero][k - one] + 1);
+          }
+        }
+      }
+    }
+    return dp[length][m][n];
+  }
+  // end::answer[]
+}