這題跟“唯一路徑”有點類似，只是多了一些障礙導致通路無法連通而已，只需要判斷一下在進行向下/向右走時是否存在障礙，分情況討論即可，而且分類討論的難度不大，也時很快就AC了。

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

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