873. Length of Longest Fibonacci Subsequence

问题：

给定一个递增数组，求其中所包含的斐波那契数列的长度。

A[i]+A[i+1]=A[i+2]

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++.)

　　

解法：

遍历最开始的两个加数，b：i=1～size，a：j=0～i-1

得出之和a+b，再在数列中查找是否存在这样的数，

递归，使得b=a+b，a=原来的b，继续求和，直到找不到，得到到目前为止的数列长度l

最终结果res=max(res,l)

代码参考：

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