LeetCode: 873. Length of Longest Fibonacci Subsequence

题目描述

A sequence X_1, X_2, …, X_n is fibonacci-like if:

• n >= 3
• X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
• Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.

(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, without changing the order of the remaining elements. For example, [3, 5, 8] is a subsequence of [3, 4, 5, 6, 7, 8].)

Example 1:

Input: [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: [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].

Note:

3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)

解题思路

枚举所有的可能性，求出最大长度。

AC 代码

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        set<int> aset(A.begin(), A.end());
        int ans = 0;
        //枚举起始两个数字
        for(int i = 0; i < A.size(); ++i)
        {
            for(int j = i+1; j < A.size(); ++j)
            {
                int cnt = 2;
                int nums[] = {A[i], A[j]};
                int cur = 0;
                while(aset.find(nums[0]+nums[1]) != aset.end())
                {
                    nums[cur] = nums[0]+nums[1];
                    cur = !cur;
                    ++cnt;
                }

                ans = max(ans, cnt);
            }
        }

        if(ans < 3) ans = 0;
        return ans;
    }
};