【leetcode】1027. Longest Arithmetic Sequence

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:

2 <= A.length <= 2000

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

巴特西

【leetcode】1027. Longest Arithmetic Sequence

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