[WXM] LeetCode 474. Ones and Zeroes C++

474. 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: 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".

Approach

題大意はあなたに一定数のゼロと1をあげて、それからあなたにArray配列の中で最も多い要素が何個あるかを詰めることができて、繰り返すことができません.この問題は416と変式の問題なので、簡単に考えられます.Partition Equal Subset Sumタイプはほぼ同じですが、ここでは2次元ですが、その問題は1次元ですが、考え方は同じです.私たちはどのようにプッシュしますか.私たちはdp[i][j] i ，j dp[i][j] を使っているので、プッシュ式 x ，y dp[i][j]=max(dp[i][j],dp[i+x][y+j]+1) dp[i-x][j-y]=max(dp[i][j]+1,dp[i-x][j-y])を出すことができます.残りはコードを見て理解を深めることができます.

[WXM] LeetCode 416. Partition Equal Subset Sum C++この問題はこの問題と似ている

Code

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        int N = strs.size();
        vectorint, int>>res(N);
        for (int i = 0; i < N; i++) {
            int one = 0, zero = 0;
            for (int j = 0; j < strs[i].size(); j++) {
                if (strs[i][j] == '0')zero++;
                else one++;
            }
            res[i] = make_pair(zero, one);
        }
        vector<vector<int>>dp(m + 1, vector<int>(n + 1, -1));
        dp[m][n] = 0;
        int maxn = 0;
        for (int i = 0; i < N; i++) {
            for (int j = res[i].first; j <= m; j++) {
                for (int k = res[i].second; k <= n; k++) {
                    if (dp[j][k] >= 0) {
                        int nj = j - res[i].first, nk = k - res[i].second;
                        dp[nj][nk] = max(dp[j][k] + 1, dp[nj][nk]);
                        maxn = max(maxn, dp[nj][nk]);
                    }
                }
            }
        }
        return maxn;
    }
};