LeetCode 1027. Longest Arithmetic Sequence--笔试题--C++解法

本文介绍了解决LeetCode 1027题——最长等差数列的方法，通过C++代码实现，探讨了如何找到数组中最长的等差子序列。示例包括输入[3,6,9,12]，输出4；输入[9,4,7,2,10]，输出3；输入[20,1,15,3,10,5,8]，输出4。

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LeetCode 1027. Longest Arithmetic Sequence–笔试题–C++解法

LeetCode题解专栏：LeetCode题解
我做的所有的LeetCode的题目都放在这个专栏里，大部分题目Java和Python的解法都有。

题目地址：Longest Arithmetic Sequence - LeetCode

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:

2 <= A.length <= 2000
0 <= A[i] <= 10000

这道题目是9月16号的XX笔试题，XX的题目把这道题改简单了。

C++解法如下：

class Solution {
public:
    int longestArithSeqLength(vector<int> &A) {
        unordered_map<int, unordered_map<int, int>> dp;
        int res = 2, n = A.size();
        for (int i = 0; i < n; ++i)
            for (int j = i + 1; j < n; ++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;
    }
};