Leetcode 474. Ones and Zeroes 01组合 解题报告

1 解题思想

这道题的意思是给了m个1和n个0，还有一堆01构成的字符串，问能用这m个1和n个0能完整组成那堆字符串里的几个字符串。
这道题的基本思想也就是DP，其实就是顺序扫描字符串，没加入一个字符串，都用DP公式计算下，DP直接看代码，不解释了

2 原题

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:The given numbers of 0s and 1s will both not exceed 100The size of given string array won't exceed 600.Example 1:Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3Output: 4Explanation: 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 = 1Output: 2Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".Subscribe to see which companies asked this question

3 AC解

public class Solution {    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表示以当前的能用的0和1 的组合，能获得的最大情况                     * */                    dp[i][j] = Math.max(1 + dp[i-count[0]][j-count[1]], dp[i][j]);        }        return dp[m][n];    }    /**     * 统计0和1的数量     * */    public int[] count(String str) {        int[] res = new int[2];        for (int i=0;i<str.length();i++)            res[str.charAt(i) - '0']++;        return res;     }}