Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
- Right -> Right -> Down -> Down
- Down -> Down -> Right -> Right
/**
* obstacleGrid = 0,0,0
* 0,1,0
* 0,0,0
* index of dp 0,1,2,3
* first time 0,1,1,1
* sec time 0,1,0,1
* third time 0,1,1,2
*
* ******************
* obstacleGrid = 0,0,0
* 0,0,0
* 0,0,0
* index of dp 0,1,2,3
* first time 0,1,1,1
* sec time 0,1,2,3
* third time 0,1,3,6
*
* @param obstacleGrid
* @return
*/
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[] dp = new int[n+1];
dp[1] = 1;
for(int i = 0; i < m; i++){
for(int j = 1; j <= n; j++){
if(obstacleGrid[i][j-1] == 1){
dp[j] = 0;
}else{
dp[j] += dp[j-1];
}
}
}
return dp[n];
}
}