Unique Paths II

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 1 and 0 respectively in the grid.

答案
在Unique Paths的基础上，只需要当遇到obstacle的时候将dp[i][j]设为0即可

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        dp[m - 1][n - 1] = 1;
        for(int i = m - 1; i >= 0; i--) {
            for(int j = n - 1; j >= 0; j--) {
                int right_paths = 0, down_paths = 0;
                // Obstacle grid
                if(obstacleGrid[i][j] == 1) {
                    dp[i][j] = 0;
                    continue;
                }
                // Skip finish grid
                if(i == m - 1 && j == n - 1) continue;
                // right
                if(j < n - 1)
                    right_paths = dp[i][j + 1];
                // down
                if(i < m - 1)
                    down_paths = dp[i + 1][j];
                dp[i][j] = right_paths + down_paths;
            }
        }
        return dp[0][0];
    }
}