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+# [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)
+
+# C++
+``` C++
+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[1][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] = 0; // 此处为障碍物
+                else dp[i][j] += (dp[i - 1][j] + dp[i][j - 1]);
+            }
+        return dp[m][n];
+    }
+};
+```