# # 【LeetCode】873. Length of Longest Fibonacci Subsequence 解题报告（Python）

## # 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++.)

## # 解题方法

``````class Solution(object):
def lenLongestFibSubseq(self, A):
"""
:type A: List[int]
:rtype: int
"""
s = set(A)
n = len(A)
res = 0
for i in range(n - 1):
for j in range(i + 1, n):
a, b = A[i], A[j]
count = 2
while a + b in s:
a, b = b, a + b
count += 1
res = max(res, count)
return res if res > 2 else 0
``````

DP.

dp[j][k] = dp[i][j] + 1 条件是 A[i] + A[j] = A[k]。

``````class Solution(object):
def lenLongestFibSubseq(self, A):
"""
:type A: List[int]
:rtype: int
"""
n = len(A)
m = dict()
for i, a in enumerate(A):
m[a] = i
res = 0
# dp[i][j] := max len of seq ends with A[i], A[j]
dp = [[2 for i in range(n)] for j in range(n)]
for j in range(n):
for k in range(j + 1, n):
a_i = A[k] - A[j]
if a_i >= A[j]:
break
if a_i in m:
i = m[a_i]
dp[j][k] = dp[i][j] + 1
res = max(res, dp[j][k])
return res
``````