【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 };