【leetcode】1027. Longest Arithmetic Sequence

标签：enum   eps   turn   self   object   下标   字典   nbsp   input   

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

【leetcode】1027. Longest Arithmetic Sequence

标签：enum   eps   turn   self   object   下标   字典   nbsp   input   

原文地址：https://www.cnblogs.com/seyjs/p/10765630.html