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.

一刷
题解：
也是DP问题，Unique Path一样可以in place解决。要点是在设置第一行和第一列碰到obstacle的时候，要将其以及之后的所有值设置为零，因为没有路径可以达到。之后在DP扫描矩阵的时候，也要讲obstacle所在的位置清零。
Time Complexity - O(m * n)， Space Complexity - O(mn)。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        
         for(int i=0; i<m; i++){
             if(obstacleGrid[i][0] == 1){
                 for(int j=i; j<m; j++){
                     dp[j][0] = 0;
                 }
                 break;
             }
             else dp[i][0] = 1;
         }
        
        for(int i=0; i<n; i++){
             if(obstacleGrid[0][i] == 1){
                 for(int j=i; j<n; j++){
                     dp[0][j] = 0;
                 }
                 break;
             }
             else dp[0][i] = 1;
         }
        
        for(int i=1; i<m; i++){
            for(int j=1; j<n; j++){
                if(obstacleGrid[i][j] == 1) dp[i][j] = 0;
                else{
                    if(i == 0||j==0) dp[i][j] = 1;
                    else dp[i][j] = dp[i-1][j] + dp[i][j-1];
                }
            }
        }
        
        return dp[m-1][n-1];
    }
}