# [63. Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)
# 思路
这题其实和[62. Unique Paths](https://leetcode.com/problems/unique-paths/)几乎一样的,唯一区别就是矩形网格里存在一些障碍物,我们对每个格子判断一下就行了。
就是一个简单的动归,可参考[62题题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/62.%20Unique%20Paths.md)。     
时间空间复杂度都是O(mn)     
另外,类似62题,可将空间复杂度优化到O(m),这里不再赘述,可参考[讨论区](https://leetcode.com/problems/unique-paths-ii/discuss/23252/4ms-O(n)-DP-Solution-in-C%2B%2B-with-Explanations)         
注意: dp数组开成int型时有的case过不了(之前是没有这些case的,直接int就能过),所以开成long long型的

# C++
``` C++
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<vector<long long> > dp(m + 1, vector<long long> (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 int(dp[m][n]);
    } 
};
```