原题链接在这里：https://leetcode.com/problems/longest-arithmetic-sequence/

题目：

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

题解：

State, let dp[diff][i] denotes with diff, up to index i, longest arithmetic length.

For j, check each i<j, with the same diff, dp[diff][j] = dp[diff][i]+1.

Time Complexity: O(n^2). n = A.length.

Space: O(n*(max-min)). max is largest value in A, min is minimum value in A. since count of distinct diff can't be more max-min.

AC Java:

 1 class Solution {
 2     public int longestArithSeqLength(int[] A) {
 3         if(A == null || A.length == 0){
 4             return 0;
 5         }
 6         
 7         int n = A.length;
 8         int res = 1;
 9         HashMap<Integer, Integer> [] dp = new HashMap[n];
10         for(int j = 0; j<n; j++){
11             dp[j] = new HashMap<>();
12             for(int i = j-1; i>=0; i--){
13                 int d = A[j] - A[i];
14                 int cur = dp[j].getOrDefault(d, 1);
15                 dp[j].put(d, Math.max(cur, dp[i].getOrDefault(d, 1)+1));
16                 res = Math.max(res, dp[j].get(d));
17             }
18         }
19         
20         return res;
21     }
22 }