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.
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31 |
public class Solution { /**This is a simple implementation of dynamic programming methodology. We use int[][] minGrid to denote the * minimum path sum from cell (0,0) to (i,j). * For each cell (i,j), there are two ways to reach it: * 1. go one step vertically from cell (i - 1 ,j) * 2. go one step horizontally from cell (i, j - 1). * 3. So the minimum path sum from cell (0,0) to (i,j) is * minGrid[i][j] = min{grid[i][j] + minGrid[i - 1][j], grid[i][j] + minGrid[i][j - 1]}; * @param grid * @return */ public
int minPathSum(int[][] grid) { int
min = 0; int
row = grid.length; if(row > 0){ int
column = grid[0].length; int[][] minGrid = new
int[row][column]; minGrid[0][0] = grid[0][0]; for(int
i = 1; i < row; ++i) minGrid[i][0] = grid[i][0]+ minGrid[i- 1][0]; for(int
i = 1; i < column; ++i) minGrid[0][i] = grid[0][i] + minGrid[0][i - 1]; for(int
i = 1; i < row; ++i){ for(int
j = 1; j < column; ++j) minGrid[i][j] = Math.min(grid[i][j] + minGrid[i - 1][j], grid[i][j] + minGrid[i][j - 1]); } min = minGrid[row - 1][column - 1]; } return
min; }} |
leetcode--Minimum Path Sum,布布扣,bubuko.com
原文地址：http://www.cnblogs.com/averillzheng/p/3773723.html
