• Difficulty: Hard
• Related Topics: Array, Union Find
• Link: https://leetcode.com/problems/longest-consecutive-sequence/

Description

Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.
給定一個無序整數陣列 nums，尋找其最長連續序列，返回其長度。

Follow up

Could you implement the O(n)

Examples

Example 1

Input: nums = [100,4,200,1,3,2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.

Example 2

Input: nums = [0,3,7,2,5,8,4,6,0,1]
Output: 9

Constraints

• 0 <= nums.length <= 1e4
• -1e9<= nums[i] <= 1e9

Solution

這題的一個比較暴力的解法就是一個個遍歷。對於每個 num，依次考察 num + 1, num + 2 等是否在陣列內，並更新結果。這種做法時間複雜度為 \(O(N^3)\)，通過將陣列轉換成 set，可以降低至 \(O(N^2)\)，程式碼如下：

import kotlin.math.max

class Solution {
    fun longestConsecutive(nums: IntArray): Int {
        if (nums.size < 2) {
            return nums.size
        }
        
        val numSet = nums.toSet()
        var result = 1
        for (num in nums) {
            var curResult = 1
            var n = num + 1
            while (n in numSet) {
                n++
                curResult++
            }
            result = max(result, curResult)
        }
        return result
    }
}
 

\(O(N)\) 的做法在 discussion 裡找到一個。建立一個 map，key 為 num，value 為 num 所在的連續序列的最大長度。對每個不在 map 裡的 num，先檢查 num + 1 和 num - 1 是否在這個 map 裡，更新 num 所在的連續序列的最大長度並儲存。程式碼如下：

import kotlin.math.max

class Solution {
    fun longestConsecutive(nums: IntArray): Int {
        var result = 0
        val map = hashMapOf<Int, Int>()

        for (num in nums) {
            if (!map.containsKey(num)) {
                val left = map[num - 1]?:0
                val right = map[num + 1]?:0

                val sum = left + right + 1
                map[num] = sum

                result = max(result, sum)

                map[num - left] = sum
                map[num + right] = sum
            }
        }
        return result
    }
}