Unique Binary Search Trees
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使用动态规划,见Leetcode题解分析,时间O(n^2),空间O(n)。代码如下:
class Solution {public: int numTrees(int n) { vector<int> f(n+1, 0); f[0] = 1; f[1] = 1; for(int i=2; i<=n; ++i) { for(int j=0; j<i; ++j) { f[i] += f[j]*f[i-j-1]; } } return f[n]; }};
9.02最新代码:
class Solution {public: int numTrees(int n) { vector<int> f(n+1); f[0] = 1; for(int i=1; i<=n; i++) { f[i] = 0; for(int j=1; j<=i; j++) { f[i] += (f[j-1]*f[i-j]); } } return f[n]; }};
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- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
- Unique Binary Search Trees
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