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.
简单的动态规划问题
class Solution { public: int dp[102][102]; int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { memset(dp,0,sizeof(dp)); dp[0][0]=obstacleGrid[0][0] == 1 ? 0:1; int n = obstacleGrid.size(), m = obstacleGrid[0].size(); for(int i = 1; i < n; ++ i ) dp[i][0] = obstacleGrid[i][0] == 1 ? 0 : dp[i-1][0]; for(int i = 1; i < m; ++ i ) dp[0][i] = obstacleGrid[0][i] == 1 ? 0 : dp[0][i-1]; for(int i = 1; i < n ; ++ i){ for(int j = 1; j < m ; ++ j){ dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i-1][j] + dp[i][j-1]; } } return dp[n-1][m-1]; } };