LeetCode474. Ones and Zeroes——动态规划

一、题目

You are given an array of binary strings strs and two integers m and n.

Return the size of the largest subset of strs such that there are at most m 0’s and n 1’s in the subset.

A set x is a subset of a set y if all elements of x are also elements of y.

Example 1:

Input: strs = [“10”,“0001”,“111001”,“1”,“0”], m = 5, n = 3
Output: 4
Explanation: The largest subset with at most 5 0’s and 3 1’s is {“10”, “0001”, “1”, “0”}, so the answer is 4.
Other valid but smaller subsets include {“0001”, “1”} and {“10”, “1”, “0”}.
{“111001”} is an invalid subset because it contains 4 1’s, greater than the maximum of 3.
Example 2:

Input: strs = [“10”,“0”,“1”], m = 1, n = 1
Output: 2
Explanation: The largest subset is {“0”, “1”}, so the answer is 2.

Constraints:

1 <= strs.length <= 600
1 <= strs[i].length <= 100
strs[i] consists only of digits ‘0’ and ‘1’.
1 <= m, n <= 100

二、题解

class Solution {
public:
    int zero = 0,one = 0;
    void zeroAndOnes(string s){
        int n = s.size();
        zero = 0;
        one = 0;
        for(int i = 0;i < n;i++){
            if(s[i] == '0') zero++;
            else one++;
        }
    }
    int findMaxForm(vector<string>& strs, int m, int n) {
        int len = strs.size();
        vector<vector<vector<int>>> dp(len + 1,vector<vector<int>>(m+1,vector<int>(n+1,0)));
        for(int i = len - 1;i >= 0;i--){
            zeroAndOnes(strs[i]);
            for(int z = 0;z <= m;z++){
                for(int o = 0;o <= n;o++){
                    int t1 = dp[i+1][z][o];
                    int t2 = 0;
                    if(zero <= z && one <= o) t2 = 1 + dp[i+1][z-zero][o-one];
                    dp[i][z][o] = max(t1,t2);
                }
            }
        }
        return dp[0][m][n];
    }
};
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