1027. Longest Arithmetic Subsequence (Solution 2)

package LeetCode_1027

import kotlin.collections.HashMap

/**
 * 1027. Longest Arithmetic Subsequence
 * https://leetcode.com/problems/longest-arithmetic-subsequence/description/
 *
 * 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: A = [3,6,9,12]
Output: 4
Explanation:
The whole array is an arithmetic sequence with steps of length = 3.

Example 2:
Input: A = [9,4,7,2,10]
Output: 3
Explanation:
The longest arithmetic subsequence is [4,7,10].

Example 3:
Input: A = [20,1,15,3,10,5,8]
Output: 4
Explanation:
The longest arithmetic subsequence is [20,15,10,5].

Constraints:
1. 2 <= A.length <= 1000
2. 0 <= A[i] <= 500
 * */
class Solution {
    /*
    * solution 1: bruce force, 3 loop, Time complexity:O(n^3), Space complexity:O(1)
    * solution 2: Array + HashMap, Time complexity:O(n^2), Space complexity:O(n)
    * */
    fun longestArithSeqLength(A: IntArray): Int {
        if (A.isEmpty()) {
            return 0
        }
        var max = 0
        val n = A.size
        //solution 2:
        /*
        * map's key is : the different of two number
        * map's value is : the count of element in Arithmetic Subsequence that has same different
        * */
        val maps = Array<HashMap<Int, Int>>(n,{ HashMap() })
        for (i in 0 until n) {
            maps[i] = HashMap()
            for (j in 0 until i) {
                val diff = A[j] - A[i]
                val mapi = maps[i]
                val mapj = maps[j]
                val currentMax = mapj.getOrDefault(diff, 1) + 1
                mapi.put(diff, currentMax)
                max = Math.max(max, currentMax)
            }
        }
        return max
    }
}

 

posted @ 2020-10-08 16:53  johnny_zhao  阅读(115)  评论(0编辑  收藏  举报