题解 | 矩阵的最小路径和

#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param matrix int整型vector<vector<>> the matrix
     * @return int整型
     */
    int minPathSum(vector<vector<int> >& matrix) {
        // write code here
        int m = matrix.size();
        int n = matrix[0].size();
        vector<vector<int>> dp(m, vector<int> (n, 0));

        dp[0][0] = matrix[0][0];
        for (int i = 1; i < m; ++i) {
            dp[i][0] = dp[i-1][0] + matrix[i][0];
        }
        for (int i = 1; i < n; ++i) {
            dp[0][i] = dp[0][i-1] + matrix[0][i];
        }

        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                dp[i][j] = matrix[i][j] + min(dp[i-1][j], dp[i][j-1]);
                // cout << dp[i][j] << " ";
            }
            // cout << endl;
        }

        return dp[m-1][n-1];
    }
};

和不同路径数目类似

Day2