1027. Longest Arithmetic Sequence

问题描述：

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:

1. 2 <= A.length <= 2000
2. 0 <= A[i] <= 10000

解题思路：

在数组中寻找最长的等差数列并且返回长度。

对于每一个数字，存在一个以该数字结尾的数列，可能只包含它一个，也可能包含其他的值。

我们对于每一个数值A[i]，遍历之前的值A[j]，看看之前的值有没有差为A[i] - A[j] 等差数列存在

如果有，我们就存入A[i]差为A[i] - A[j]的等差数列的长度为之前的长度加一。

这里我用了两层map嵌套实现

外层：<index, innerMap>

内层：<diff, length>

我从下标为一的值开始向后遍历并且检索前面的，并没有初始化map，所以要对不存在的键值进行处理。

代码：

class Solution {
public:
    int longestArithSeqLength(vector<int>& A) {
        //we can using a map to store the possible arithmetic sequence.
        //the key is the index of array
        //the value is the map of possible difference of arithmetic sequence and the length of its
        //we can use a variable to record current maximum length
        //time complexity is O(n^2)
        if(A.size() < 3) return A.size();  
        unordered_map<int, unordered_map<int,int> > m;
        int n = A.size();
        int ret = 1;
        for(int i = 1; i < n; ++i){
            for(int j = 0; j < i; ++j){
                int diff = A[i]-A[j];
                if(m.count(j) != 0 && m[j].count(diff) != 0){
                    m[i][diff] = m[j][diff];
                }else{
                    m[i][diff] = 1;
                }
                m[i][diff]++;
                ret = max(ret, m[i][diff]);
                // std::cout << A[i] << " : " << A[j] << std::endl;
                // std::cout << diff << " : " << ret << std::endl;
            }
        }
        //++ret;
        return ret;
    }
};