首页 > 其他 > 详细

LintCode Minimum Path Sum

时间:2016-08-11 12:55:25      阅读:176      评论:0      收藏:0      [点我收藏+]

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.

 Notice

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

Dynamic programming is ultilized to solve this problem.

First of all, define another matrix which has same dimension for both x and y. And let us define the number stored in the array stand for the minimum summation of all the path to the position with same x and y in grid matrix. 

Then the sum[i][j] = min(sum[i-1][j],sum[i][j-1]) + grid[i][j];

Initialize the sum[0][0] = grid [0][0]; initialize the two boundaries with all the summation of previous path to that certain node.

Solve sum[m-1][n-1]

 

 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         // write your code here
 8         if (grid == null || grid.length ==0 || grid[0].length == 0) {
 9             return 0;
10         }
11         int m = grid.length;
12         int n = grid[0].length;
13         for(int i = 1; i < m; i++) {
14             grid[i][0] = grid[i-1][0] + grid[i][0];
15         }
16         for(int i = 1; i < n; i++) {
17             grid[0][i] = grid[0][i-1] + grid[0][i];
18         }
19         for(int i= 1; i < m; i ++) {
20             for (int j =1; j < n; j++) {
21                 grid[i][j] = Math.min(grid[i-1][j],grid[i][j-1]) + grid[i][j];
22             }
23         }
24         return grid[m-1][n-1];
25     }
26 }

 

 

LintCode Minimum Path Sum

原文:http://www.cnblogs.com/ly91417/p/5760450.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!