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 0sand 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 mostonce.

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 = 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".

二维01背包问题。res[i][j]=max(res[i][j],1+res[i-zero][j-one]) 前者当前字符串不放入背包，后者将字符串放入背包。

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