Minimum Path Sum

最新推荐文章于 2019-10-26 15:10:44 发布

转载最新推荐文章于 2019-10-26 15:10:44 发布 · 37 阅读

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原文链接：http://www.cnblogs.com/beiyeqingteng/p/5652094.html

本文介绍了一个算法问题：在一个填充了非负数的矩阵中，从左上角到右下角寻找一条路径，使得路径上的数字之和最小。文中提供了一个Java实现方案，通过动态规划的方法初始化矩阵的第一行和第一列，并逐步计算出每个位置上的最小路径和。

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.

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

 1 public class Solution {
 2     /**
 3      * @param grid: a list of lists of integers.
 4      * @return: An integer, minimizes the sum of all numbers along its path
 5      */
 6     public int minPathSum(int[][] grid) {
 7         if (grid == null) return 0;
 8         
 9         // initialize the first row
10         int [][] min = new int[grid.length][grid[0].length];
11         
12         min[0][0] = grid[0][0];
13         
14         for (int i = 1; i < grid[0].length; i++) {
15             min[0][i] = min[0][i - 1] + grid[0][i];
16         }
17         
18         for (int i = 1; i < grid.length; i++) {
19             min[i][0] = min[i - 1][0] + grid[i][0];
20         }
21         
22         for (int i = 1; i < grid.length; i++) {
23             for (int j = 1; j < grid[0].length; j++) {
24                 min[i][j] = (int)Math.min(min[i - 1][j], min[i][j - 1]) + grid[i][j];
25             }
26         }
27         
28         return min[grid.length - 1][grid[0].length - 1];
29     }
30 }

转载于:https://www.cnblogs.com/beiyeqingteng/p/5652094.html