LeetCode Unique Binary Search Trees

Unique Binary Search Trees

Givenn, how many structurally uniqueBST's(binary search trees) that store values 1...n?

For example,
Givenn= 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

动态规划法

本题关键：如何填表，如何得到计算公式.

需要很仔细画图，一步一步找出规律。

Catlan公式是可以计算这样的题目的，不过Catalan公式也不好理解，还是写个循环解决吧。

class Solution {
public:
	int numTrees(int n)
	{
		vector<int> ta(n+1);
		ta[0] = 1; ta[1] = 1; ta[2] = 2;
		for (int i = 3; i <= n; i++)
		{
			int mid = (i-1)/2;

			for (int j = 0; j < mid; j++)
			{
				ta[i] += ta[j] * ta[i-j-1] *2;
			}

			if (i%2) ta[i] += ta[mid] * ta[mid];
			else ta[i] += ta[mid+1] * ta[mid] * 2;
		}
		return ta[n];
	}
};

//2014-2-15 update
	int numTrees(int n)
	{
		if (n < 3) return n;
		int *A = new int[n+1];
		A[0] = 1, A[1] = 1, A[2] = 2;
		for (int i = 3; i <= n; i++)
		{
			int half = i/2;
			A[i] = 0;//注意要初始化
			for (int j = 1; j <= half; j++)
				A[i] += A[j-1] * A[i-j] *2;//注意细节是+=
			if (i%2 == 1) A[i] += A[half]*A[half];
		}
		return A[n];
	}
	//2014-2-15 update 2
	int numTrees2(int n)
	{
		int *A = new int[n+1];
		A[0] = 1;
		for (int i = 1; i <= n; i++)
		{
			A[i] = 0;//注意要初始化
			for (int j = 1; j <= i; j++)
				A[i] += A[j-1] * A[i-j];//注意细节是+=
		}
		return A[n];
	}

下面是参考资料：

http://zh.wikipedia.org/wiki/%E5%8D%A1%E5%A1%94%E5%85%B0%E6%95%B0

http://en.wikipedia.org/wiki/Catalan_number