[LeetCode] Length of Longest Fibonacci Subsequence

 

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++.)

 使用set来存储数组中的元素，然后选取两个元素遍历set。找出符合a+b的结果出现的值，随后更新a、b的值继续遍历set。选择符合条件的最大次数即可。

class Solution:
    def lenLongestFibSubseq(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        s = set(A)
        cnt = 0
        for i in range(len(A)):
            for j in range(i+1, len(A)):
                a, b = A[i], A[j]
                t = 2
                while a+b in s:
                    a, b = b, a+b
                    t = t+1
                if t >= 3:
                    cnt = max(cnt, t)
        return cnt