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.

 

 1 public class Solution {
 2     public int minPathSum(int[][] grid) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         //[m,n] = (m,n) + min([m-1,n],[m,n-1])
 6         //only can move right or down
 7         if(grid == null || grid.length == 0 || grid[0].length == 0)return 0;
 8         int m = grid.length;
 9         int n = grid[0].length;
10         int[][] map = new int[m][n];
11         for(int i = 0; i < m; i ++){
12             map[i][0] = grid[i][0];
13             if(i > 0) map[i][0] += map[i - 1][0];
14         }
15         for(int j = 1; j < n; j ++){
16             map[0][j] = map[0][j - 1] + grid[0][j];
17         }
18         for(int i = 1; i < m; i ++){
19             for(int j = 1; j < n; j ++){
20                 map[i][j] = grid[i][j] + Math.min(map[i - 1][j], map[i][j - 1]);
21             }
22         }
23         return map[m - 1][n - 1];
24     }
25 }

 第二遍：

 1 public class Solution {
 2     public int minPathSum(int[][] grid) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         //[m,n] = (m,n) + min([m-1,n],[m,n-1])
 6         //only can move right or down
 7         if(grid == null || grid.length == 0 || grid[0].length == 0)return 0;
 8         int m = grid.length;
 9         int n = grid[0].length;
10         int[][] map = new int[m][n];
11         for(int i = 0; i < m; i ++){
12             for(int j = 0; j < n; j ++){
13                 map[i][j] = grid[i][j];
14                 if(i > 0 && j > 0) map[i][j] += Math.min(map[i - 1][j], map[i][j - 1]);
15                 else if(i > 0) map[i][j] += map[i - 1][j];
16                 else if(j > 0) map[i][j] += map[i][j - 1];
17             }
18         }
19         return map[m - 1][n - 1];
20     }
21 }

 第三遍：

 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, n = grid[0].length;
 5         int[][] result = new int[m][n];
 6         result[0][0] = grid[0][0];
 7         for(int i = 0; i < m; i ++)
 8             for(int j = 0; j < n; j ++)
 9                 if(i != 0 || j != 0){
10                     int aa = i > 0 ? result[i - 1][j] : Integer.MAX_VALUE;
11                     int bb = j > 0 ? result[i][j - 1] : Integer.MAX_VALUE;
12                     result[i][j] = Math.min(aa, bb) + grid[i][j];
13                 }
14         return result[m - 1][n - 1];
15     }
16 }

 

posted on 2013-10-03 03:16  Step-BY-Step  阅读(199)  评论(0)  编辑  收藏  举报

导航