[leetcode] 1027. Longest Arithmetic Subsequence

Description

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

Example 2:

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

Example 3:

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

Constraints:

• 2 <= A.length <= 1000
• 0 <= A[i] <= 500

分析

题目的意思是：找出最长算术子序列，即子序列两数之间的差为定长。这道题的解法也很巧妙，首先是动态规划，但是dp数组里面存放的是字典，字典的键为差值，值为差值的统计量，这一点是要明确的。
接下来就是计算求出最大长度，对于当前的A[i]，计算前面的数A[j]与A[i]的差值，然后更新这个dp数组，维护最大长度res就新行了。递推公式为：

dp[i][diff]=max(2,dp[j][diff]+1)

根据题目的意思，等差数列最小长度为2，所以就从2开始了哈哈。

代码

class Solution:
    def longestArithSeqLength(self, A: List[int]) -> int:
        n=len(A)
        d=[defaultdict(int) for i in range(n)]
        res=2
        for i in range(n):
            for j in range(i):
                diff=A[i]-A[j]
                d[i][diff]=max(2,d[j][diff]+1)
                res=max(res,d[i][diff])
        return res

参考文献

[LeetCode] Python, DP O(n^2) Solution