474. Ones and Zeroes

本周继续练习动态规划的相关题目。

题目：

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.

示例：

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".


题解：
　　同样地，这道题需要通过动态规划来解决，因为不能贪心地去换尽量短的字符,当两个字符长度相同的时候，用贪心算法无法决定选哪一个。
　　因此，这道题需要通过动态规划来解决。题目中给定m个0与n个1，dp[m][n]就标记了在当前的字符串序列下，能够用m个0与n个1取得的最多字符串。
　　显然，dp[0][0] = 0,因为没有0与1可用的时候，必然得不到字符串
　　　　dp[m][n] = max{dp[m][n],dp[m-m0][n-n0]+1}
　　针对每一个字符串，实质上我们都采取两种策略，选这个字符串，还是不选。做决定之前，先统计这个字符串(第i个)有多少个0与1，分别记为m0与n0，如果不选这个字符串，那么dp[m][n]的值不会发生任何改变，但是选了之后，它的值就是用了m-m0个0 与n-n0个1，所得到的最多的字符串，也就是f[m-m0][n-n0]再加上1，有点类似于经典0-1背包问题
　　
代码：

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) { 
        // m个0 n个1
        vector<vector<int> > dp(m+1,vector<int>(n+1,0));
        for (int i = 0 ; i < strs.size();i++)
        {
            string curr = strs[i];
            int num0 = 0,num1 = 0;
            for(int j = 0; j < curr.length();j++)
            {
                if(curr[j] == '0') num0++;
                if(curr[j] == '1') num1++;
            }
            
            for(int j = m; j >=num0; j--)
            {
                for(int i = n;i >= num1 ;i--)
                {
                    dp[j][i] = max(dp[j][i],dp[j-num0][i-num1]+1);
                }
            }
            
        }
        
        return dp[m][n];
    }
};