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: 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 题目大意： 这道题目在62题的基础上添加了障碍物，求解从左上角到右下角所有的路径之和。   解法： 同样使用动态递归法，在初始化第一排和第一列的值的时候，如果遇到障碍物，然么这一排和这一列后面都为0。

class Solution {
private:
    double dp[105][105];
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m=obstacleGrid.size(),n=obstacleGrid[0].size();
        if(m==0||n==0) return 0;
        for(int i=0;i<m;i++){
            if(!obstacleGrid[i][0]) dp[i][0]=1;
            else break;
        }
        for(int i=0;i<n;i++){
            if(!obstacleGrid[0][i]) dp[0][i]=1;
            else break;
        }
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                if(!obstacleGrid[i][j]) dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }

        return dp[m-1][n-1];
    }
};