[Swift]LeetCode1027. 最长等差数列 | Longest Arithmetic Sequence

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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

---

给定一个整数数组 A，返回 A 中最长等差子序列的长度。

回想一下，A 的子序列是列表 A[i_1], A[i_2], ..., A[i_k] 其中 0 <= i_1 < i_2 < ... < i_k <= A.length - 1。并且如果 B[i+1] - B[i]( 0 <= i < B.length - 1) 的值都相同，那么序列 B 是等差的。

示例 1：

输入：[3,6,9,12]
输出：4
解释： 
整个数组是公差为 3 的等差数列。

示例 2：

输入：[9,4,7,2,10]
输出：3
解释：
最长的等差子序列是 [4,7,10]。

示例 3：

输入：[20,1,15,3,10,5,8]
输出：4
解释：
最长的等差子序列是 [20,15,10,5]。

提示：

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

---

Runtime: 1424 ms

Memory Usage: 75.9 MB

class Solution {
    func longestArithSeqLength(_ A: [Int]) -> Int {
        var n:Int = A.count
        var map:[[Int:Int]] = [[Int:Int]](repeating:[Int:Int](),count:n)
        var longest:Int = 1
        for i in 1..<n
        {
            for j in 0..<i
            {
                var d:Int = A[i] - A[j]
                var l:Int = map[j][d,default:1] + 1
                map[i][d] = max(l, map[i][d,default:1])
                longest = max(longest, l)
            }
        }
        return longest       
    }
}

---

2688ms

class Solution {
    func longestArithSeqLength(_ A: [Int]) -> Int {
        if A.count <= 2{
            return A.count
        }
        
        var dp = [Int:[Int:Int]]()
        var longest = 0
        for base in 0..<A.count{
            let baseVal = A[base]
            for current in base+1..<A.count{
                let currentVal = A[current]
                let diff = currentVal - baseVal
                
                dp[current, default: [Int:Int]()][diff] = max(
                                     dp[base, default: [Int:Int]()][diff, default: 1]+1, 
                                     dp[current, default: [Int:Int]()][diff, default: 1])
                
                longest = max(longest, dp[current]![diff]!)
            }
        }
        
        return longest
    }
}