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25 lines
1.3 KiB
Markdown
25 lines
1.3 KiB
Markdown
# [63. Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)
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# 思路
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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。
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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)。
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时间空间复杂度都是O(mn)
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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)
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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
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# C++
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``` C++
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class Solution {
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public:
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int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
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int m = obstacleGrid.size(), n = obstacleGrid[0].size();
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vector<vector<long long> > dp(m + 1, vector<long long> (n + 1, 0));
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dp[0][1] = 1;
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for (int i = 1; i <= m; i++)
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for (int j = 1; j <= n; j++)
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if (!obstacleGrid[i - 1][j - 1])
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
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return int(dp[m][n]);
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}
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};
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```
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