[LeetCode]Unique Binary Search Trees, 解题报告
2013-12-05 11:11
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前言
今天准备把LeetCode上AC Rates在30%以上的题目全部收尾,这道题目也是我在这个AC Rate上遇到的感觉最麻烦的一道题目,分享一下自己的解题思路而且这道题目AC Rate竟然这么高
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.
思路
这道题目我想了两天,一直试图找出递推的规律,昨晚才想明白思路有问题,这应该是一道典型的动态规划题目,因为有重叠的子问题,当前决策依赖于子问题的解我设dp[i]表示共有i个节点时,能产生的BST树的个数
n == 0 时,空树的个数必然为1,因此dp[0] = 1
n == 1 时,只有1这个根节点,数量也为1,因此dp[1] = 1
n == 2时,有两种构造方法,如下图所示:
因此,dp[2] = dp[0] * dp[1] + dp[1] * dp[0]
n == 3时,构造方法如题目给的示例所示,dp[3] = dp[0] * dp[2] + dp[1] * dp[1] + dp[2] * dp[0]
同时,当根节点元素为 1, 2, 3, 4, 5, ..., i, ..., n时,基于以下原则的BST树具有唯一性:
以i为根节点时,其左子树构成为[0,...,i-1],其右子树构成为[i+1,...,n]构成
因此,dp[i] = sigma(dp[0...k] * dp[k+1...i]) 0 <= k < i - 1
代码
import java.util.Scanner; public class UniqueBinarySearchTrees { public static int numTrees(int n) { int i, j, dp[] = new int[n + 1]; if (n == 0) return 1; dp[0] = dp[1] = 1; for (i = 2; i <= n; i++) { dp[i] = 0; for (j = 0; j < i; j++) { dp[i] += dp[j] * dp[i - j - 1]; } } return dp ; } public static void main(String[] args) { int n, num; Scanner cin = new Scanner(System.in); while (cin.hasNext()) { n = cin.nextInt(); num = numTrees(n); System.out.println(num); } cin.close(); } }
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