[LeetCode]Length of Longest Fibonacci Subsequence

题目描述：

LeetCode 873. 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:

<strong>Input: </strong>[1,2,3,4,5,6,7,8]
<strong>Output: </strong>5
<strong>Explanation:
</strong>The longest subsequence that is fibonacci-like: [1,2,3,5,8].

Example 2:

<strong>Input: </strong>[1,3,7,11,12,14,18]
<strong>Output: </strong>3
<strong>Explanation</strong>:
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++.)

题目大意：

求递增正整数组的最长类斐波那契数列。

解题思路：

动态规划（Dynamic Programming）

时间复杂度 O(N^2)

状态转移方程：

dp[y][x + y] = max(dp[y][x + y], dp[x][y] + 1)

上式中, dp[x][y]表示以(x, y)为结尾的类斐波那契数列的长度

Python代码：

class Solution(object):
    def lenLongestFibSubseq(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        vset = set(A)
        dp = collections.defaultdict(lambda: collections.defaultdict(int))
        size = len(A)
        ans = 0
        for i in range(size):
            x = A[i]
            for j in range(i + 1, size):
                y = A[j]
                if x + y not in vset: continue
                dp[y][x + y] = max(dp[y][x + y], dp[x][y] + 1)
                ans = max(dp[y][x + y], ans)
        return ans and ans + 2 or 0