leetcode problem 96. Unique Binary Search Trees

最新推荐文章于 2021-03-31 10:21:52 发布

原创最新推荐文章于 2021-03-31 10:21:52 发布 · 135 阅读

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探讨了如何计算给定整数n时，可以构造的不同结构的二叉搜索树的数量。通过递推公式fn=f0×fn−1+f1×fn−2+⋯+fn−1×f0，使用C++代码实现了解决方案，展示了运行时间和内存使用的优化策略。

Description

Given n, how many structurally unique BST’s (binary search trees) that store values 1 … n?

Example:

Input: 3
Output: 5
Explanation:
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

Solution

好像是个什么数列：
$fn=f0×fn−1+f1×fn−2+⋯+fn−1×f0f_n= f_0 \times f_{n-1} +f_1 \times f_{n-2} + \dots +f_{n-1} \times f_0$

class Solution {
public:
    int numTrees(int n) {
        vector<int> res(n+1,0);
        res[0] = 1;
        res[1] = 1;
        for(int i = 2; i <= n; ++i)
            for(int j = 0; j < i; ++j)
                res[i] += res[j] * res[i-1-j];
        return res[n];
    }
};

result:

Runtime: 4 ms, faster than 100.00% of C++ online submissions for Unique Binary Search Trees.
Memory Usage: 8.3 MB, less than 34.53% of C++ online submissions for Unique Binary Search Trees.

好像不能同时快且内存小啊。

20190415

tbc