题目如下:

Given an array A of integers, return the length of the longest arithmetic subsequence in A.

Recall that a subsequence of A is a list A[i_1], A[i_2], ..., A[i_k] with 0 <= i_1 < i_2 < ... < i_k <= A.length - 1, and that a sequence B is arithmetic if B[i+1] - B[i] are all the same value (for 0 <= i < B.length - 1).

Example 1:

Input: [3,6,9,12]
Output: 4
Explanation:
The whole array is an arithmetic sequence with steps of length = 3.

Example 2:

Input: [9,4,7,2,10]
Output: 3
Explanation:
The longest arithmetic subsequence is [4,7,10].

Example 3:

Input: [20,1,15,3,10,5,8]
Output: 4
Explanation:
The longest arithmetic subsequence is [20,15,10,5].

Note:

  1. 2 <= A.length <= 2000
  2. 0 <= A[i] <= 10000

解题思路:首先用字典记录A中每个元素出现的下标,接下来求出任意A[i]与A[j]的差值d,依次判断A[j] += d是否存在于A中,并且要求A[j] + d的下标的最小值要大于j,最终即可求出最长的等差数列。

代码如下:

class Solution(object):
def longestArithSeqLength(self, A):
"""
:type A: List[int]
:rtype: int
"""
import bisect
res = 0
dic = {}
for i,v in enumerate(A):
dic[v] = dic.setdefault(v,[]) + [i]
for i in range(len(A)):
for j in range(i+1,len(A)):
count = 2
diff = A[j] - A[i]
next = A[j] + diff
smallestInx = j
while True:
if next not in dic:
break
inx = bisect.bisect_right(dic[next],smallestInx)
if inx == len(dic[next]):
break
smallestInx = dic[next][inx]
next = next + diff
count += 1
res = max(res,count)
return res

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