[leetcode] 873. Length of Longest Fibonacci Subsequence

Description

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:

1. 3 <= A.length <= 1000
2. 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++.)

分析

题目的意思是：题目的意思是给你一个数组，找出最长的fibonacci数列，我开始用到暴力求解的方法，写了两个大循环，后面看答案还有动态规划的解法，我这里欣赏一下，顺便分享出来，方便大家一同进步。大致思路是首先建立一个关于A的值和索引的字典，然后按照求最长递增子序列的方法遍历，如果(i,j)，(j,k)相连，则

longest[j, k] = longest[i, j] + 1

细节我也不是很懂，经常看应该就懂了

代码

class Solution:
    def lenLongestFibSubseq(self, A: List[int]) -> int:
        index_dict={}
        for index,item in enumerate(A):
            index_dict[item]=index
        longest=collections.defaultdict(lambda:2)
        
        res=0
        for index,value in enumerate(A):
            for j in range(index):
                i=index_dict.get(value-A[j],None)
                if(i is not None and i<j):
                    longest[j,index]=longest[i,j]+1
                    res=max(longest[j,index],res)
        if(res<3):
            res=0
        return res 
        

参考文献

​​[LeetCode] Approach 2: Dynamic Programming​​