LeetCode 96:Unique Binary Search Trees

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

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//由1,2,3,...,n構建的二叉查找樹，以i為根節點，左子樹由[1,i-1]構成，其右子樹由[i+1,n]構成。
//定義f(i)為以[1,i]能產生的Unique Binary Search Tree的數目
//若數組為空，則只有一種BST，即空樹，f(0)=1;
//若數組僅有一個元素1，則只有一種BST，單個節點，f(1)=1;
//若數組有兩個元素1，2，則有兩種可能，f(2)=f(0)*f(1)+f(1)*f(0);
//若數組有三個元素1，2，3，則有f(3)=f(0)*f(2)+f(1)*f(1)+f(2)*f(0)
//由此可以得到遞推公式：f(i)=f(0)*f(i-1)+...+f(k-1)*f(i-k)+...+f(i-1)*f(0)
//利用一維動態規劃來求解
class Solution {
public:
	int numTrees(int n) {
		vector f(n+1,0); //n+1個int型元素，每個都初始化為0
		f[0] = 1;
		f[1] = 1;
		for (int i = 2; i <= n; ++i){
			for (int k = 1; k <= i; ++k)
				f[i] = f[i] + f[k - 1] * f[i - k];
		}
		return f[n];
	}
};