第十一周：[Leetcode]474. Ones and Zeroes-优快云博客

本文介绍了一个计算机科学问题：如何利用有限的0和1资源形成给定字符串数组中的最多字符串数量。通过动态规划的方法解决了这一问题，并给出了具体的实现代码。

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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:
The given numbers of 0s and 1s will both not exceed 100
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”.

动态规划：dp[i][j]表示i个0，j个1最多能组成多少个给定数组中的字符串，假如数组中一个字符串为str，str中0的个数为zeros，1的个数为ones，则状态转移方程为dp[i][j] = max(dp[i-zeros][j-ones] + 1,dp[i][j])，从后往前更新dp数组。

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