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LeetCode: Unique Paths II 解题报告

时间:2015-04-08 06:37:46      阅读:111      评论:0      收藏:0      [点我收藏+]

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Unique Paths II

 Total Accepted: 31019 Total Submissions: 110866My Submissions

 

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

 

SOULUTION 1:

LeetCode: Unique Paths 解题报告 相比,只不过是判断一下当前是不是block,如果是block,直接设置D[i][j] 是0就好了 也就是不可达。 同样在3分钟之内Bug free 秒答,哦耶!

技术分享
 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         // 1150
 4         if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) {
 5             return 0;
 6         }
 7         
 8         int rows = obstacleGrid.length;
 9         int cols = obstacleGrid[0].length;
10         
11         int[][] D = new int[rows][cols];
12         
13         for (int i = 0; i < rows; i++) {
14             for (int j = 0; j < cols; j++) {
15                 D[i][j] = 0;
16                 if (obstacleGrid[i][j] == 1) {
17                     D[i][j] = 0;
18                 } else {
19                     if (i == 0 && j == 0) {
20                         D[i][j] = 1;
21                     }
22                     
23                     if (i != 0) {
24                         D[i][j] += D[i - 1][j];
25                     }
26                     
27                     if (j != 0) {
28                         D[i][j] += D[i][j - 1];
29                     }
30                 }
31             }
32         }
33         
34         return D[rows - 1][cols - 1];
35     }
36 }
View Code

 

LeetCode: Unique Paths II 解题报告

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原文地址:http://www.cnblogs.com/yuzhangcmu/p/4401268.html

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