【leetcode】Unique Paths II

Unique Paths II

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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.

把存在Obstacle的位置标记为-1，表示无法通行
 
 
 1 class Solution {
 2 public:
 3     int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
 4         int m=obstacleGrid.size();
 5         int n=obstacleGrid[0].size();
 6         
 7         int dp[101][101];
 8         
 9         dp[0][0]=obstacleGrid[0][0]==1?-1:1;
10         
11         for(int i=1;i<m;i++)
12         {
13             if(obstacleGrid[i][0]==1) 
14             {
15                 dp[i][0]=-1;
16                 continue;
17             }
18             if(dp[i-1][0]==-1) dp[i][0]=-1;
19             else dp[i][0]=1;
20         }
21         
22         for(int j=1;j<n;j++)
23         {
24             if(obstacleGrid[0][j]==1) 
25             {
26                 dp[0][j]=-1;
27                 continue;
28             }
29             
30             if(dp[0][j-1]==-1) dp[0][j]=-1;
31             else dp[0][j]=1;
32         }
33         
34         for(int i=1;i<m;i++)
35         {
36             for(int j=1;j<n;j++)
37             {
38                 if(obstacleGrid[i][j]==1) 
39                 {
40                     dp[i][j]=-1;
41                     continue;
42                 }
43                 if(dp[i-1][j]==-1&&dp[i][j-1]==-1) dp[i][j]=-1;
44                 else if(dp[i-1][j]==-1&&dp[i][j-1]!=-1) dp[i][j]=dp[i][j-1];
45                 else if(dp[i-1][j]!=-1&&dp[i][j-1]==-1) dp[i][j]=dp[i-1][j];
46                 else if(dp[i-1][j]!=-1&&dp[i][j-1]!=-1) dp[i][j]=dp[i-1][j]+dp[i][j-1];
47             }
48         }
49         
50         return dp[m-1][n-1]==-1?0:dp[m-1][n-1];
51     }
52 };

 

 
 
 
实际上不用标记也可以，dp[i][j]=0就表示了没有路
 
 1 class Solution {
 2 public:
 3     int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
 4         int m=obstacleGrid.size();
 5         int n=obstacleGrid[0].size();
 6         
 7         int dp[101][101];
 8         //vector<vector<int>> dp(m,vector<int>(n));
 9         
10         dp[0][0]=obstacleGrid[0][0]==1?0:1;
11         for(int i=1;i<m;i++)
12         {
13             dp[i][0]=obstacleGrid[i][0]==1?0:dp[i-1][0];
14         }
15         for(int j=1;j<n;j++)
16         {
17             dp[0][j]=obstacleGrid[0][j]==1?0:dp[0][j-1];
18         }
19         for(int i=1;i<m;i++)
20         {
21             for(int j=1;j<n;j++)
22             {
23                 dp[i][j]=obstacleGrid[i][j]==1?0:dp[i-1][j]+dp[i][j-1];
24             }
25         }
26         
27         return dp[m-1][n-1];
28     }
29 };

 

posted @ 2015-01-05 14:27  H5开发技术  阅读(163)  评论(0)  编辑  收藏  举报