[LeetCode每日一题]63. Unique Paths II

题目如下：

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
Now consider if some obstacles are added to the grids. How many unique paths would there be?

Example:
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:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

63. Unique Paths II
这道题是在62. Unique Paths（解法见[LeetCode每日一题]62. Unique Paths）的基础上修改的，添加了障碍物，设状态空间为dp[][],状态转移方程为

dp[i][j] = dp[i-1][j] + dp[i][j-1],  obstacleGrid[i][j]==0
dp[i][j] = 0,  obstacleGrid[i][j]==1

实现代码如下：

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid[0] == null) {
            return 0;
        }
        int x = obstacleGrid.length;
        int y = obstacleGrid[0].length;
        int[][] dp = new int[x][y];
        dp[0][0] = obstacleGrid[0][0] == 0 ? 1 : 0;
        for (int i = 1; i < x; i++) {
            dp[i][0] = obstacleGrid[i][0] == 1 ? 0 : dp[i - 1][0];
        }
        for (int j = 1; j < y; j++) {
            dp[0][j] = obstacleGrid[0][j] == 1 ? 0 : dp[0][j - 1];
        }
        for (int i = 1; i < x; i++) {
            for (int j = 1; j < y; j++) {
                dp[i][j] = obstacleGrid[i][j] == 0 ? dp[i - 1][j] + dp[i][j - 1] : 0;
            }
        }
        return dp[x - 1][y - 1];
    }
}