[LeetCode]Unique Binary Search Trees
2016-11-15 19:28
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Question
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.
本题难度Medium。
DP
【复杂度】
时间 O(N) 空间 O(N)
【思路】
可以参照(复习)[LeetCode]Unique Binary Search Trees II,设n的BST的个数为
f(1,n)=∑i=1nf(1,i−1)∗f(i+1,n)
实际上,
f(i+1,n)=f(1,n−i)
因此可以进一步地化简,我们用
f(n)=∑i=1nf(i−1)∗f(n−i)
然后我们利用DP对
【代码】
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
本题难度Medium。
DP
【复杂度】
时间 O(N) 空间 O(N)
【思路】
可以参照(复习)[LeetCode]Unique Binary Search Trees II,设n的BST的个数为
f(1,n),则可以看到:
f(1,n)=∑i=1nf(1,i−1)∗f(i+1,n)
实际上,
f(i+1,n)=f(1,n−i)
因此可以进一步地化简,我们用
f(n)表示BST的个数,则:
f(n)=∑i=1nf(i−1)∗f(n−i)
然后我们利用DP对
f(0)到
f(n)的值依次进行计算即可。
【代码】
public class Solution { public int numTrees(int n) { //require int[] f=new int[n+1]; if(n<1) f=new int[2]; f[0]=1;f[1]=1; //invariant for(int k=2;k<=n;k++) for(int i=1;i<=k;i++) f[k]+=f[i-1]*f[k-i]; //ensure return f ; } }
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