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Update 63. Unique Paths II.md

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唐树森 2019-02-18 23:28:44 +08:00 committed by GitHub
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solutions/63. Unique Paths II.md
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@ -3,6 +3,8 @@
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++
@ -10,14 +12,13 @@ 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;
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] = 0; // 此处为障碍物
else dp[i][j] += (dp[i - 1][j] + dp[i][j - 1]);
}
return dp[m][n];
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]);
}
};
```