LeetCode题目：Unique Binary Search Trees，一维动态规划

转载自：http://blog.unieagle.net/2012/11/01/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Aunique-binary-search-trees%EF%BC%8C%E4%B8%80%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/

用一维动态规划解。
Sigma(左边的子树可能状态 * 右边子树可能状态) = 当前个数的nodes形成的BST总数。

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

代码：8ms过大集合

class Solution {public:    int numTrees(int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(n <= 1) return n;        //ways[i] rep. the number of ways with i nodes        int *ways = new int[n + 1];        ways[0] = 1;        ways[1] = 1;        for(int i = 2 ; i <= n; ++i) {            ways[i] = 0;            for(int left = 0; left < i; ++ left) {                //with number of left noeds in the left sub-tree                int lways = ways[left];                int rways = ways[i - left - 1];                ways[i] += lways * rways;            }        }        int ret = ways[n];        delete [] ways;        return ret;    }};