Add to List 474. Ones and Zeroes（第十五周）

Description：

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".

解题思路：

这是背包一类问题，从后往前更新，才能得到答案。

#include <iostream>#include <vector>#include <cstring>using namespace std;class Solution {public:    int findMaxForm(vector<string>& strs, int m, int n) {        int size = strs.size();        int dp[m+1][n+1];        memset(dp,0,sizeof(dp));    //    int countNum[size][2];        for(int i = 0; i < size; i++)        {        int count0 = 0, count1 = 0;         for(int j = 0; j < strs[i].size(); j++)        {        if(strs[i][j] == '0') count0++;        if(strs[i][j] == '1') count1++;}//countNum[i][0] = count0;//countNum[j][1] = count1;for(int j = m; j >= count0; j--){for(int k = n; k >= count1; k--){dp[j][k] = max(dp[j][k], dp[j-count0][k-count1]+1);}}}return dp[m][n];    }};