474. Ones and Zeroes

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 0and 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".

DP求解，定义一个二维数组，int[][] dp = new int[m+1][n+1]，其中dp[i][j]表示i个0和j个1能表示字符串的最大个数，假设当前字符串有a个0，b个1，则有dp[i][j] = Math.max(dp[i][j], dp[i-a][j-b])，因此程序如下所示：

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m+1][n+1];
        for (String s : strs){
            int zeros = 0, ones = 0;
            int len = s.length();
            for (int i = 0; i < len; ++ i){
                if (s.charAt(i) == '0'){
                    zeros ++;
                }
                else {
                    ones ++;
                }
            }
            for (int i = m; i >= zeros; -- i){
                for (int j = n; j >= ones; -- j){
                    dp[i][j] = Math.max(dp[i][j], dp[i-zeros][j-ones] + 1);
                }
            }
        }
        return dp[m][n];
    }
}