105 lines
2.6 KiB
Markdown
105 lines
2.6 KiB
Markdown
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## 题目地址
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https://leetcode-cn.com/problems/longest-arithmetic-subsequence-of-given-difference/
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## 题目描述
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```
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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。
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示例 1：
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输入：arr = [1,2,3,4], difference = 1
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输出：4
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解释：最长的等差子序列是 [1,2,3,4]。
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示例 2：
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输入：arr = [1,3,5,7], difference = 1
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输出：1
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解释：最长的等差子序列是任意单个元素。
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示例 3：
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输入：arr = [1,5,7,8,5,3,4,2,1], difference = -2
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输出：4
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解释：最长的等差子序列是 [7,5,3,1]。
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提示：
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1 <= arr.length <= 10^5
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-10^4 <= arr[i], difference <= 10^4
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```
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## 思路
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最直观的思路是双层循环，我们暴力的枚举出以每一个元素为开始元素，以最后元素结尾的的所有情况。很明显这是所有的情况，这就是暴力法的精髓， 很明显这种解法会TLE（超时），不过我们先来看一下代码，顺着这个思维继续思考。
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### 暴力法
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```python
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def longestSubsequence(self, arr: List[int], difference: int) -> int:
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n = len(arr)
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res = 1
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for i in range(n):
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count = 1
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for j in range(i + 1, n):
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if arr[i] + difference * count == arr[j]:
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count += 1
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if count > res:
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res = count
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return res
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```
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### 动态规划
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上面的时间复杂度是O(n^2)， 有没有办法降低到O(n)呢？很容易想到的是空间换时间的解决方案。
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我的想法是将`以每一个元素结尾的最长等差子序列的长度`统统存起来，即`dp[num] = maxLen` 这样我们遍历到一个新的元素的时候，就去之前的存储中去找`dp[num - difference]`, 如果找到了，就更新当前的`dp[num] = dp[num - difference] + 1`, 否则就是不进行操作（还是默认值1）。
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这种空间换时间的做法的时间和空间复杂度都是O(n)。
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## 关键点解析
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- 将`以每一个元素结尾的最长等差子序列的长度`统统存起来
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## 代码
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```python
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#
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# @lc app=leetcode.cn id=1218 lang=python3
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#
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# [1218] 最长定差子序列
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#
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# @lc code=start
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class Solution:
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# 动态规划
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def longestSubsequence(self, arr: List[int], difference: int) -> int:
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n = len(arr)
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res = 1
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dp = {}
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for num in arr:
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dp[num] = 1
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if num - difference in dp:
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dp[num] = dp[num - difference] + 1
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return max(dp.values())
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# @lc code=end
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```
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## 相关题目
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