96. Unique Binary Search Trees
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96. Unique Binary Search Trees
题目:
Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
For example :
Given n = 3, there are a total of 5 unique BST's. 1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3
开始时,尝试用递归实现,但是超时了,可见系统对运行时间有要求。因为递归过程中存在大量的重复计算,从n一层层往下递归,故考虑类似于动态规划的思想,让底层的计算结果能够被重复利用,故用一个数组存储中间计算结果(即 1~n-1 对应的BST数目),这样只需双层循环即可.
代码如下:
public int numTrees(int n) { int [] G = new int[n+1]; G[0] = G[1] = 1; for(int i=2; i<=n; ++i) { for(int j=1; j<=i; ++j) { G[i] += G[j-1] * G[i-j]; } } return G[n];}
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- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
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