Minimum Path Sum

最新推荐文章于 2019-10-26 15:10:44 发布

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最新推荐文章于 2019-10-26 15:10:44 发布

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原文链接：http://www.cnblogs.com/luckygxf/p/4106761.html

本文详细解析了如何使用动态规划解决最小路径和问题，通过一个m x n的网格寻找从左上角到右下角的路径，使得路径上的数字之和最小。文中提供了一个具体的Java代码实现，并解释了递推关系及边界条件。

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.

用DP实现

找出递推关系和边界条件

 1 public class Solution {
 2     public int minPathSum(int[][] grid) {
 3         if(1 == grid.length && 1 == grid[0].length)
 4             return grid[0][0];
 5         int result[][] = new int[grid.length + 1][grid[0].length + 1];
 6         
 7             
 8             for(int i = 1; i <= grid.length; i++){
 9                 for(int j = 1; j <= grid[0].length; j++){
10                     if(i == 1){
11                         result[i][j] = result[i][j - 1]; 
12                     }
13                     else if(j == 1)
14                         result[i][j] = result[i - 1][j];
15                     else
16                         result[i][j] = result[i - 1][j] <= result[i][j - 1] ? result[i - 1][j] : result[i][j - 1];
17                     result[i][j] += grid[i - 1][j - 1];
18                 }
19             }
20 
21             return result[grid.length][grid[0].length];
22     }
23 }