[题目]
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.
[分析] n个节点的BST的结构数等于从1到n依次为root节点对应的BST个数,记录中间结果避免重复计算。
public class Solution { // Method 2 //dp[i] : number of unique BSTs which store value [1,i] public int numTrees(int n) { if (n <= 0) return 0; int[] dp = new int[n + 1]; dp[0] = 1; for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) // select j as root dp[i] += dp[j - 1] * dp[i - j]; } return dp[n]; } // Method 1 public int numTrees(int n) { if (n <= 0) return 0; int[] dp = new int[n + 1]; dp[0] = 1; dp[1] = 1; for (int i = 2; i <= n; i++) { int mid = (i - 1) / 2; for (int j = 0; j <= mid; j++) dp[i] += 2 * dp[j] * dp[i - 1 - j];//编码过程中i-1-j写成n-1-j,调试了好一会 if (i % 2 == 1) dp[i] -= dp[mid] * dp[mid]; } return dp[n]; } }