[LeetCode] 128. Longest Consecutive Sequence（最长连续序列）

• 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.
给定一个无序整数数组 nums，寻找其最长连续序列，返回其长度。

Follow up

Could you implement the O(n) solution?
你能以 \(O(N)\) 时间复杂度解决吗？

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

这题的一个比较暴力的解法就是一个个遍历。对于每个 num，依次考察 num + 1, num + 2 等是否在数组内，并更新结果。这种做法时间复杂度为 \(O(N^3)\)，通过将数组转换成 set，可以降低至 \(O(N^2)\)，代码如下：

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 里找到一个。建立一个 map，key 为 num，value 为 num 所在的连续序列的最大长度。对每个不在 map 里的 num，先检查 num + 1 和 num - 1 是否在这个 map 里，更新 num 所在的连续序列的最大长度并保存。代码如下：

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
    }
}