Unique Paths II

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.

这道题目是Unique Paths的变形题目，用1代表障碍物，有障碍物的地方就是死路，同样采用动态规划的思想，只是多了一些限定条件，当为1时，我们就把当前点的值设定为0；当为0的时候，如果此时（i = 0或者j = 0）这时当前点的值等于它前面的点的值（相当于初始化的过程），如果此时（i > 0并且j > 0)，动态转移方程为dp[i][j] = dp[i - 1][j] + dp[i][j - 1],(i >= 1, j >= 1)。代码如下：

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