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
discussion: https://leetcode.com/problems/longest-arithmetic-sequence/discuss/274611/JavaC++Python-DP
思路：
dp：unordered_map<int, unordered_map<int, int>> dp来记录, dp[diff][i]，差为diff的以i为结尾的等差序列长度。 initialize: map不需要initialize，但是要注意等差序列的最短长度也是2，所以每一次更新和2取max。 transfer: dp[diff][i] = dp[diff].count(i) ? dp[diff][i] + 1 : 2。也就是说，对于两个位置上 j < i的两个数字，首先我们一定会将这个 (i, j) 对记录下来，也就是说至少dp[diff][i] = 2。但是如果在之前 j 位置上已经存在dp[diff][i]，甚至可能大于2，那么就要在其基础上累加1，取代2，成为在 j 位置上以diff为差值的最长等差序列长度。 return：每一次循环中同步更新的全球最大值。
Time complexity: O(n^2) Space complexity: O(n)
int longestArithSeqLength(vector<int>& A) { unordered_map<int, unordered_map<int, int>> dp; int res = 2; for (int i = 0; i < A.size(); ++i) for (int j = i + 1; j < A.size(); ++j) { int d = A[j] - A[i]; dp[d][j] = dp[d].count(i) ? dp[d][i] + 1 : 2; res = max(res, dp[d][j]); } return res; }