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LeetCode873.LengthofLongestFibonacciSubsequence(2025/2/2

LeetCode873.LengthofLongestFibonacciSubsequence(2025/2/2

昨天工作耽搁了,没来得及打卡每日一题,今日补上:

标题:Length of Longest Fibonacci Subsequence

题目:

例子:

Example 1:

Input: arr = [1,2,3,4,5,6,7,8] Output: 5 Explanation: The longest subsequence that is fibonacci-like: [1,2,3,5,8].

Example 2:

Input: arr = [1,3,7,11,12,14,18] Output: 3 Explanation: The longest subsequence that is fibonacci-like: [1,11,12], [3,11,14] or [7,11,18].

解题思路:

先看这道题数量级是10的3次方,因此时间复杂度能接受O(n^2),但不能接受O(n^3)。确定解题用到的算法及数据结构:要求最长子序列的长度,只要确定子序列最前两个数字,后面的数字都是确定的,因此两层循环确定最前面两个数字,查看整个序列有多少个数字包含在序列中即可。查看的过程可以用unordered_set,查询时间复杂度为O(1)。

代码:

class Solution { public: int lenLongestFibSubseq(vector<int>& arr) { unordered_set<int> set(arr.begin(), arr.end()); int res = 0; for(int i = 0; i < arr.size() - 2; i++){ for(int j = i+1; j < arr.size() - 1; j ++){ int len = 2, start = arr[i], second = arr[j], next = arr[i] + arr[j]; while(next <= arr.back() && set.count(next)) { len ++; start = second; second = next; next = start + second; } if (len > 2) res = max(res, len); } } return res; } };

O(N^2)<时间复杂度<O(N^3)

空间复杂度为O(N)

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