[LeetCode] 121: Unique Binary Search Trees

[Problem]

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

[Solution]
方法一
class Solution {
public:
int numTrees(int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.

// invalid
if(n <= 0)return 0;

// initial
int dp[n+1];
memset(dp, 0, sizeof(dp));
dp[0] = 1;
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= i; ++j){
dp[i] += dp[j-1]*dp[i-j];
}
}
return dp[n];
}
};
方法二:Catalan数 C_n+1= C_n * 2(2n+1)/(n+2)

class Solution {
public:
int numTrees(int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.

// invalid
if(n <= 0)return 0;

// initial
int dp[n+1];
memset(dp, 0, sizeof(dp));
dp[1] = 1;
for(int i = 2; i <= n; ++i){
dp[i] = 2*(2*(i-1)+1)*dp[i-1]/(i-1+2);
}
return dp[n];
}
};


说明:版权所有,转载请注明出处。 Coder007的博客
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值