题解 | #字符串的排列#

动态规划。

matrix

1 3 5 9

8 1 3 4

5 0 6 1

8 8 4 0

dp

1 4 9 18

9 5 8 12

14 5 11 12

22 13 15 12

public int minPathSum (int[][] matrix) {
        // write code here
        int m = matrix.length;
        int n = matrix[0].length;

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

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = matrix[i][j] + Math.min(dp[i-1][j], dp[i][j-1]);
            }
        }
        return dp[m-1][n-1];
    }