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.

n个数中任何一个数都可以当成根节点，左边j个结点，右边n-j-1个

所以就有nums[n] = nums[0]*nums[n - 1]+nums[1]*nums[n-2]+nums[2]*nums[n-3]+...+nums[n-1]*nums[0]; 因此我们需要先求出nums[0]--nums[n-1] 最后返回nums[n]即可

 

class Solution {

    public int numTrees(int n) {

        int[] arr = new int[n + 1];

        arr[0] = arr[1] = 1;

        for (int i = 2; i <= n; i++) {

            for (int j = 0; j < i; j++)

                arr[i] += arr[j] * arr[i - j - 1];

        }   

        return arr[n];

    }

}