474.Ones and Zeroes

最新推荐文章于 2022-06-30 23:01:26 发布

原创最新推荐文章于 2022-06-30 23:01:26 发布 · 338 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#string #leetcode-java

LeetCode 专栏收录该内容

137 篇文章

订阅专栏

本文针对LeetCode上的474题 Ones and Zeroes 提供了解题思路及Java实现代码。该问题属于0-1背包问题的变种，通过动态规划的方法求解在给定0和1数量限制下能组成的最大字符串数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

474.Ones and Zeroes

题目描述：In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.

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.

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.

Note:
1. The given numbers of 0s and 1s will both not exceed 100
2. The size of given string array won’t exceed 600.

Example 1:

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”

Example 2:

Input: Array = {"10", "0", "1"}, m = 1, n = 1
Output: 2

Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".

题目大意：给定一个只包含0与1的字符数组array，给定m表述0最多出现的次数，给定n表示1最多出现的次数，问在给定的m与n的范围内，可以从array中选出的最多字符的个数。
思路：DP，0,1背包的变种，dp[i][j]表示m为i,n为j能找出的最多字符串的个数。

代码

package DP;

/**
* @author OovEver
* 2017/12/25 9:59
*/
public class LeetCode474 {
  public int findMaxForm(String[] strs, int m, int n) {
      int[][] dp = new int[m + 1][n + 1];
      for (String s : strs) {
          int[] count = count(s);
          for(int i=m;i>=count[0];i--) {
              for(int j=n;j>=count[1];j--) {
                  dp[i][j] = Math.max(dp[i][j], 1 + dp[i - count[0]][j - count[1]]);
              }
          }
      }
      return dp[m][n];
  }

  /**
   * 统计s中0与1的数量
   * @param s 要统计的字符串
   * @return 字符串中0与1的数量
   */
  private int[] count(String s) {
      int[] res = new int[2];
      for(int i=0;i<s.length();i++) {
          res[s.charAt(i) - '0']++;
      }
      return res;
  }
}