leetcode96~Unique Binary Search Trees
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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
动态规划问题。
此问题的关键:1 BST,左边的树都比根小,右边的树都比根大。因此,k为根节点时,左边的树为1..k-1,右边的树为k+1…n
2 以某个节点为根节点的BST可能性=左子树BST的可能性*右子叔BST的可能性(乘法组合)
2 n个数,每个节点都可能成为根节点,累加起来即可。count[i]表示i个节点的bst数量。
求解:G(n)表示n个节点数时的bst数量 F(i,n)表示以i为根节点的bst数量,i=1....n G(n)=F(1,n)+F(2,n)+...F(n,n) G(0)=G(1)=1 F(i,n)=G(i-1)*G(n-i) i=1....n ===>G(n) = G(0)*G(n-1)+G(1)*G(n-2)+…+G(n-1)*G(0)
public class UniqueBST { public int numTree(int n) { int[] G = new int[n+1]; G[0]=G[1]=1; //从节点数2开始计算到n for(int i=2;i<=n;i++) { for(int j=0;j<i;j++) { G[i] = G[i]+G[j]*G[i-1-j]; } } return G[n]; } public int numTrees1(int n) { if(n==0 || n==1) { return 1; } int[] count= new int[n+1]; count[0] = 1; for(int num=1;num<=n;num++) { for(int i=0;i<=num-1;i++) { count[num] = count[num] + count[i]*count[num-1-i]; } } return count[n]; }}
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