64. Minimum Path Sum

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.

本题和triangle比较相似,只不过从三角形变成了矩形,但是方法还是一样的,同样是用dp解决,代码如下:

 1 public class Solution {
 2     public int minPathSum(int[][] grid) {
 3         if(grid==null||grid.length==0||grid[0].length==0) return 0;
 4         int m = grid.length;
 5         int n =grid[0].length;
 6         int[][] dp = new int[m][n];
 7         dp[0][0] = grid[0][0];
 8         for(int i=1;i<m;i++){
 9             dp[i][0] = dp[i-1][0]+grid[i][0];
10         }
11         for(int i=1;i<n;i++){
12             dp[0][i] = dp[0][i-1]+grid[0][i];
13         }
14         for(int i=1;i<m;i++){
15             for(int j=1;j<n;j++){
16                 dp[i][j] =grid[i][j]+Math.min(dp[i][j-1],dp[i-1][j]);
17             }
18         }
19         return dp[m-1][n-1];
20     }
21 }
 
原文地址:https://www.cnblogs.com/codeskiller/p/6385748.html