LeetCode Unique Paths II
2014-07-14 14:07
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题目
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
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
The total number of unique paths is
Note: m and n will be at most 100.
和上一题类似,只是有些格不能走。
对于可以走的格,p(i,j)=p(i-1,j)+p(i,j-1);
对于不能走的格,p(i,j)=0;
代码:
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
1and
0respectively 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.
和上一题类似,只是有些格不能走。
对于可以走的格,p(i,j)=p(i-1,j)+p(i,j-1);
对于不能走的格,p(i,j)=0;
代码:
class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { int m=obstacleGrid.size(); if(m==0) return 0; int n=obstacleGrid[0].size(); if(n==0) return 0; vector<vector<int>> paths(m,vector<int>(n,0)); int i,j; if(obstacleGrid[0][0]==1) return 0; paths[0][0]=1; for(i=1;i<m;i++) if(obstacleGrid[i][0]==1) paths[i][0]=0; else paths[i][0]=paths[i-1][0]; for(j=1;j<n;j++) if(obstacleGrid[0][j]==1) paths[0][j]=0; else paths[0][j]=paths[0][j-1]; for(i=1;i<m;i++) for(j=1;j<n;j++) if(obstacleGrid[i][j]==1) paths[i][j]=0; else paths[i][j]=paths[i-1][j]+paths[i][j-1]; return paths[m-1][n-1]; } };
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