leetcode------Unique Paths II

标题： Unique Paths II
通过率： 28%
难度： 中等

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.

与第一个版本一样，但是原始数组中会出现1的情况那么针对1来处理，如果遇到1则将其置-1，然后在路径相加中如果是-1则不进行相加

代码如下：

 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         int m=obstacleGrid.length;
 4         int n=obstacleGrid[0].length;
 5         boolean flag=true;
 6         if(obstacleGrid[0][0]==1)return 0;
 7         for(int i=0;i<m;i++){
 8             if(obstacleGrid[i][0]!=1&&flag)
 9                 obstacleGrid[i][0]=1;
10             else {
11                 obstacleGrid[i][0]=-1;
12                 flag=false;
13             }
14         }
15         flag=true;
16         for(int i=1;i<n;i++){
17             if(obstacleGrid[0][i]!=1&&flag)
18                 obstacleGrid[0][i]=1;
19             else {
20                 obstacleGrid[0][i]=-1;
21                 flag=false;
22             }
23         }
24         for(int i=1;i<m;i++){
25             for(int j=1;j<n;j++){
26                 if(obstacleGrid[i][j]!=1){
27                     if(obstacleGrid[i-1][j]==-1&&obstacleGrid[i][j-1]==-1)
28                         obstacleGrid[i][j]=0;
29                     if(obstacleGrid[i-1][j]==-1)
30                         obstacleGrid[i][j]+=obstacleGrid[i][j-1];
31                     else if(obstacleGrid[i][j-1]==-1)
32                         obstacleGrid[i][j]+=obstacleGrid[i-1][j];
33                     else obstacleGrid[i][j]=obstacleGrid[i-1][j]+obstacleGrid[i][j-1];
34                 }
35                 else{
36                     obstacleGrid[i][j]=-1;
37                 }
38             }
39         }
40             if(obstacleGrid[m-1][n-1]==-1)return 0;
41             else 
42                 return obstacleGrid[m-1][n-1];
43         
44     }
45 }

 

posted @ 2015-03-05 11:35  pku_smile  阅读(127)  评论(0)  编辑  收藏  举报