算法描述：

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?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

Note: m and n will be at most 100.

Example 1:

Input:[  [0,0,0],  [0,1,0],  [0,0,0]]Output: 2Explanation: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 -> Down2. Down -> Down -> Right -> Right

解题思路：这道题和Unique Paths 解法类似，只不过多了一个障碍判断。需要重点关注的是初始化阶段也要进行判断，最上边一行和最左边一列中有障碍，则障碍后面的都置为0；

int uniquePathsWithObstacles(vector
     
      
       >& obstacleGrid) {        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        vector
       
        
         > dp(m,vector
         
          (n,0));        for(int i=0; i < m; i++){            if(obstacleGrid[i][0] == 1) break;            dp[i][0]=1;        }        for(int j=0; j < n; j++){            if(obstacleGrid[0][j] == 1) break;            dp[0][j]=1;        }                    for(int i = 1; i < m ; ++i)            for(int j = 1 ; j < n ; ++j){                if(obstacleGrid[i][j]==1) continue;                dp[i][j] = dp[i-1][j]+dp[i][j-1];            }        return dp[m-1][n-1];    }