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Given an integer array arr and an integer difference, return the length of the longest arithmetic subsequence in arr such that the difference between adjacent elements in the subsequence equals difference.
Example 1:
Input: arr = [1,2,3,4], difference = 1 Output: 4 Explanation: The longest arithmetic subsequence is [1,2,3,4].
Example 2:
Input: arr = [1,3,5,7], difference = 1 Output: 1 Explanation: The longest arithmetic subsequence is any single element.
Example 3:
Input: arr = [1,5,7,8,5,3,4,2,1], difference = -2 Output: 4 Explanation: The longest arithmetic subsequence is [7,5,3,1].
Constraints:
1 <= arr.length <= 10^5-10^4 <= arr[i], difference <= 10^4
给你一个整数数组 arr 和一个整数 difference，请你找出 arr 中所有相邻元素之间的差等于给定 difference 的等差子序列，并返回其中最长的等差子序列的长度。
示例 1：
输入：arr = [1,2,3,4], difference = 1 输出：4 解释：最长的等差子序列是 [1,2,3,4]。
示例 2：
输入：arr = [1,3,5,7], difference = 1 输出：1 解释：最长的等差子序列是任意单个元素。
示例 3：
输入：arr = [1,5,7,8,5,3,4,2,1], difference = -2 输出：4 解释：最长的等差子序列是 [7,5,3,1]。
提示：
1 <= arr.length <= 10^5-10^4 <= arr[i], difference <= 10^4
Runtime: 748 ms
Memory Usage: 23.9 MB
1 class Solution { 2 func longestSubsequence(_ arr: [Int], _ difference: Int) -> Int { 3 var dp:[Int:Int] = [Int:Int]() 4 var longest:Int = 0 5 for i in 0..<arr.count 6 { 7 dp[arr[i]] = dp[arr[i]-difference,default:0] + 1 8 longest = max(longest, dp[arr[i],default:0]) 9 } 10 return longest 11 } 12 }