[leetcode] 63.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.

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.
题意：这道题是根据第62道题延伸得到，这道题增加了一些障碍，如果某个位置可以通过，那么这个位置上的数字就是0，如果这个位置不能通过，那么这个位置上的数字就是1.

思路：依旧使用动态规划，只不过如果A[i][j] = 1，那么DP[i][j] = 0。其余跟第62道题一样。

1. if(A[0][0] = 0) DP[0][0] = 1; else return 0
2. if(A[i][j] == 1)DP[i][j] = 0;else DP[i][j] = (i==0? 0 : DP[i-1][j]) + (j == 0? 0 : DP[i][j-1])(!(i==0 && j== 0))

3. return DP[m][n]

   代码如下：

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        if(m == 0)return 0;        int **DP = new int*[m];        for(int i = 0; i < m; i++){            DP[i] = new int[n];        }        if(obstacleGrid[0][0] == 0)DP[0][0] = 1;        else return 0;        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(i == 0 && j == 0)continue;                if(obstacleGrid[i][j] == 1)DP[i][j] = 0;                else DP[i][j] = (i==0?0:DP[i-1][j]) + (j == 0? 0 : DP[i][j-1]);            }        }        return DP[m-1][n-1];    }};