• 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
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13781872.html
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