Leetcode - Minnum Path-优快云博客

本文详细解析了在一个非负整数矩阵中寻找从左上角到右下角的路径，使得路径上的数值之和最小的问题。采用动态规划的方法，通过递归公式计算每个节点的最小路径和，并最终返回右下角节点的最小路径和。

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

[分析] 倒推思路，从后往前推，要求走到最后一个格子的minPath, 若知道所有可能的倒数第二个格子的minPath,取其小者加上最后一格的值即可，题中之前一个只能上相邻的上面一格和左边一格，如此可得递推公式。

public class Solution {
    // dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
    public int minPathSum(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0)
            return 0;
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][n];
        dp[0][0] = grid[0][0];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i > 0 && j > 0) 
                    dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
                else if (i > 0)
                    dp[i][j] = dp[i - 1][j] + grid[i][j];
                else if (j > 0)
                    dp[i][j] = dp[i][j - 1] + grid[i][j];
            }
        }
        return dp[m - 1][n - 1];
    }
}