# 873. Length of Longest Fibonacci Subsequence

873. Length of Longest Fibonacci Subsequence

A sequence X_1, X_2, ..., X_nis 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++.)

  通过散列表可以很好的解决此题。以所给数组中的元素为key值，建立哈希表，通过双重循环遍历遍历哈希表，令a = A[i] b = A[j]看看在哈希表中能否找到以a+b为key的value值，如果能找到，则输出长度+1，通过更新a与b的值继续找，遍历完成后输出最大长度。

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        unordered_set<int> hash(A.begin(), A.end());
        int res = 0;
        int n = A.size();
        for(int i = 0; i<n; i++)
        {
            for(int j = i + 1; j<n; j++)
            {
                int a = A[i];
                int b = A[j];
                int countnums = 2;
                while(hash.count(a+b))
                {
                    b = a + b;
                    a = b - a;
                    countnums++;
                }
                res = max(res, countnums);
            }
        }
        return res > 2 ? res : 0;
    }
};