Leetcode - Minnum Path

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];
    }
}

 

评论

2 楼 likesky3 2015-04-16  

谢谢yb君指正~

1 楼 qb_2008 2015-04-15  

递推式需修正