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Problem

Given an array nums of integers, return the length of the longest arithmetic subsequence in nums.

Recall that a subsequence of an array nums is a list nums[i1], nums[i2], …, nums[ik] with 0 <= i1 < i2 < … < ik <= nums.length - 1, and that a sequence seq is arithmetic if seq[i+1] - seq[i] are all the same value (for 0 <= i < seq.length - 1).

Example 1:

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

Example 2:

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

Example 3:

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

Code

class Solution {
    public int longestArithSeqLength(int[] nums) {
        int res = 2;
        int len = nums.length;

        HashMap<Integer, Integer>[] dp = new HashMap[len];

        for (int i = 0; i < len; i++) {
            dp[i] = new HashMap<>();

            for (int j = 0; j < i; j++) {
                int diff = nums[i] - nums[j];
                // 之前的数, 具有同样的diff
                dp[i].put(diff, dp[j].getOrDefault(diff, 1) + 1);
                res = Math.max(res, dp[i].get(diff));
            }
        }

        return res;
    }
}