标签：style class blog code java http
题目：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
个人思路：
1、二叉搜索树的中序遍历是有序的（从小到大），这里是1-n，那么我就从1-n遍历根节点，将每个情况下的BST个数相加，便是结果
2、根节点左边的便是左子树，右边的便是右子树，再用相同的方法去处理左右子树，左子树的个数乘以右子树的个数便是以该节点为根节点的BST的个数，这是一个递归的过程，递归的基本条件是空树的情况返回1
代码：
1 #include <iostream> 2 3 class Solution 4 { 5 public: 6 int numTrees(int n) 7 { 8 int num = 0; 9 10 for (int i = 1; i <= n; ++i) 11 { 12 num += partNumTrees(1, i - 1) * partNumTrees(i + 1, n); 13 } 14 15 return num; 16 } 17 private: 18 int partNumTrees(int start, int end) 19 { 20 if (start > end) 21 { 22 return 1; 23 } 24 25 int num = 0; 26 for (int i = start; i <= end; ++i) 27 { 28 num += partNumTrees(start, i - 1) * partNumTrees(i + 1, end); 29 } 30 31 return num; 32 } 33 }; 34 35 int main() 36 { 37 Solution s; 38 std::cout << s.numTrees(4) << std::endl; 39 system("pause"); 40 41 return 0; 42 }
按照惯例，上网搜寻更优的算法，发现有利用动态规划思想解决该问题的算法，链接：http://www.blogjava.net/menglee/archive/2013/12/20/407801.html
思路：
1、先开辟一个n+1大小的数组dp，dp[i]记录了有i个节点的BTS的个数，这样就可以在需要时直接读取数值，而不是当场去计算，用空间换取时间
2、主要目的是要计算出dp[n]，可从dp[2]开始计算，先预设几个dp初值，如dp[0]=1,dp[1]=1等，计算dp[i]的思路与我上面的思路相同
代码：
1 #include <iostream> 2 3 class Solution 4 { 5 public: 6 int numTrees(int n) 7 { 8 /* 9 int num = 0; 10 11 for (int i = 1; i <= n; ++i) 12 { 13 num += partNumTrees(1, i - 1) * partNumTrees(i + 1, n); 14 } 15 */ 16 17 int *dp = new int[n + 1]; 18 dp[0] = 1; 19 dp[1] = 1; 20 21 for (int i = 2; i <= n; ++i) 22 { 23 int tmp = 0; 24 for (int j = 1; j <= i; ++j) 25 { 26 tmp += dp[j - 1] * dp[i - j]; 27 } 28 dp[i] = tmp; 29 } 30 31 return dp[n]; 32 } 33 private: 34 /* 35 int partNumTrees(int start, int end) 36 { 37 if (start > end) 38 { 39 return 1; 40 } 41 42 int num = 0; 43 for (int i = start; i <= end; ++i) 44 { 45 num += partNumTrees(start, i - 1) * partNumTrees(i + 1, end); 46 } 47 48 return num; 49 } 50 */ 51 }; 52 53 int main() 54 { 55 Solution s; 56 std::cout << s.numTrees(4) << std::endl; 57 system("pause"); 58 59 return 0; 60 }
leetcode - Unique Binary Search Trees,布布扣,bubuko.com
leetcode - Unique Binary Search Trees
标签：style class blog code java http
原文地址：http://www.cnblogs.com/laihaiteng/p/3789279.html
