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  1. Math

Catalan Number

PreviousMathNextCombinatorics

Last updated 3 years ago

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The nth Catalan number is given directly in terms of binomial coefficients by

Cn=1n+1(2nn)=(2n)!(n+1)!n!=∏k=2nn+kkfor n≥0C_n = \dfrac{1}{n+1}{2n \choose n} = \dfrac{(2n)!}{(n+1)!n!}=\prod_{k=2}^{n}\dfrac{n+k}{k}\quad \text{for } n \ge 0Cn​=n+11​(n2n​)=(n+1)!n!(2n)!​=k=2∏n​kn+k​for n≥0
// OJ: https://leetcode.com/problems/unique-binary-search-trees
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numTrees(int n) {
        long long ans = 1, i;
        for (i = 1; i <= n; ++i) ans = ans * (i + n) / i;
        return ans / i;
    }
};

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Reference

96. Unique Binary Search Trees (Medium)
241. Different Ways to Add Parentheses (Medium)
https://en.wikipedia.org/wiki/Catalan_number