Ones and Zeroes

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 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.

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".

解析：

动态规划，找到递推关系式：

dp[i][j]=max(dp[i][j],dp[i-zerosnum][j-onesnum]+1)

其中dp表示i个0和j个1最多能组成多少字符串，问题是每个字符串只能用一次，开始从0开始计算，这样会出现一个字符串重复利用的情况，所以要倒着计算，即从m和n开始循环计算dp[i][j].

代码：

class Solution {public:    int findMaxForm(vector<string>& strs, int m, int n) {        vector<vector<int>>dp(m+1,vector<int>(n+1,0));        dp[0][0]=0;        for (int k=0; k<strs.size(); k++)        {            int zerosnum=0;            int onesnum=0;            for (int p=0; p<strs[k].size(); p++)            {                if (strs[k][p]=='0')                {                    zerosnum++;                }                else                {                    onesnum++;                }            }                         for (int i=m; i>=zerosnum; i--)            {                for (int j=n; j>=onesnum; j--)                {                    dp[i][j]=max(dp[i-zerosnum][j-onesnum]+1,dp[i][j]);                                    }            }                    }                       return dp[m][n];    }};