Leetcode 474. 一和零

动态规划

class Solution {
public:

    vector<int> getonezeros(string &str) {
        vector<int> onezeros(2, 0);
        for (auto s : str) {
            if (s == '0')   onezeros[0] += 1;
            if (s == '1')   onezeros[1] += 1;
        }
        return onezeros;
    }

    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 = 1; i < len + 1; i++) {
            vector<int> onezero = getonezeros(strs[i - 1]);
            int zeros = onezero[0], ones = onezero[1];

            for (int j = 0; j < m + 1; j++) {
                for (int k = 0; k < n + 1; k++) {
                    if (zeros <= j && ones <= k)
                        dp[i][j][k] = max(dp[i - 1][j][k], dp[i - 1][j - zeros][k - ones] + 1);
                    else
                        dp[i][j][k] = dp[i - 1][j][k]; 
                }
            }
        } 
        return dp[len][m][n];
    }
};

滚动数组

递推公式：dp[m][n] = max(dp[m][n], dp[m - zeros][n - ones] + 1);

class Solution {
public:

    vector<int> getonezeros(string &str) {
        vector<int> onezeros(2, 0);
        for (auto s : str) {
            if (s == '0')   onezeros[0] += 1;
            if (s == '1')   onezeros[1] += 1;
        }
        return onezeros;
    }

    int findMaxForm(vector<string>& strs, int m, int n) {
        int len = strs.size();

        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));

        for (int i = 0; i < len ; i++) {
            vector<int> onezero = getonezeros(strs[i]);
            int zeros = onezero[0], ones = onezero[1];

            for (int j = m; j >= zeros; j--) {
                for (int k = n; k >= ones; k--) {
                    dp[j][k] = max(dp[j][k], dp[j - zeros][k - ones] + 1);
                }
            }
        } 
        return dp[m][n];
    }
};