标签:
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
int numTrees(int n) {
	if (n<=0) return 0;
	if (n==1) return 1;
	if (n==2) return 2;
	vector<int>a(n+1,0);
	a[0]=1;
	a[1]=1;
	a[2]=2;
	for (int i=3;i<=n;i++)
	{
		int temp=0;
		for (int j=0;j<i;j++)
		{
			temp+=a[j]*a[i-j-1];
		}
		a[i]=temp;
	}
	return a[n];
}方法二:int numTrees1(int n)
{
	if (n<=0) return 0;
	if (n==1) return 1;
	if (n==2) return 2;
	unsigned long c1=fac(2*n);
	unsigned long c=fac(n);
	int res=c1/(c*c*(n+1));
	return res;
}
unsigned long fac(int n)
{
	vector<unsigned long> a(n+1,1);
	a[0]=1;
	a[1]=1;
	for (int i=2;i<=n;i++)
	{
		a[i]=i*a[i-1];
	}
	return a[n];
}#include <iostream>
#include <vector>
using namespace std;
int numTrees(int n);
int numTrees1(int n);
unsigned long fac(int n);
void main()
{
	int n=11;
	cout<<numTrees(n)<<endl;
	for (int i=1;i<10;i++)
	{
		cout<<numTrees1(i)<<endl;//从1到9输出第二种方法的结果
	}
}
int numTrees(int n) {
	if (n<=0) return 0;
	if (n==1) return 1;
	if (n==2) return 2;
	vector<int>a(n+1,0);
	a[0]=1;
	a[1]=1;
	a[2]=2;
	for (int i=3;i<=n;i++)
	{
		int temp=0;
		for (int j=0;j<i;j++)
		{
			temp+=a[j]*a[i-j-1];
		}
		a[i]=temp;
	}
	return a[n];
}
int numTrees1(int n)
{
	if (n<=0) return 0;
	if (n==1) return 1;
	if (n==2) return 2;
	unsigned long c1=fac(2*n);
	unsigned long c=fac(n);
	int res=c1/(c*c*(n+1));
	return res;
}
unsigned long fac(int n)
{
	vector<unsigned long> a(n+1,1);
	a[0]=1;
	a[1]=1;
	for (int i=2;i<=n;i++)
	{
		a[i]=i*a[i-1];
	}
	return a[n];
}结果如下:leetcode-Unique Binary Search Trees:
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原文地址:http://blog.csdn.net/sinat_24520925/article/details/45562273