# Longest Consecutive Sequence ### Source - leetcode: [Longest Consecutive Sequence | LeetCode OJ](https://leetcode.com/problems/longest-consecutive-sequence/) - lintcode: [(124) Longest Consecutive Sequence](http://www.lintcode.com/en/problem/longest-consecutive-sequence/) ### Problem Given an unsorted array of integers, find the length of the longestconsecutive elements sequence. #### Example Given `[100, 4, 200, 1, 3, 2]`,The longest consecutive elements sequence is `[1, 2, 3, 4]`. Return itslength: `4`. #### Clarification Your algorithm should run in O(*n*) complexity. ### 题解 首先看题要求，时间复杂度为 O(n)O(n)O(n), 如果排序，基于比较的实现为 nlognn \log nnlogn, 基数排序需要数据有特征。故排序无法达到复杂度要求。接下来可以联想空间换时间的做法，其中以哈希表为代表。这个题要求返回最长连续序列，不要求有序，非常符合哈希表的用法。**由于给定一个数其连续的数要么比它小1，要么大1，那么我们只需往左往右搜索知道在数组中找不到数为止。**结合哈希表查找为 O(1)O(1)O(1) 的特性即可满足要求。 ### Java ~~~ public class Solution { /** * @param nums: A list of integers * @return an integer */ public int longestConsecutive(int[] num) { if (num == null || num.length == 0) return 0; // add number to hashset Set<Integer> hashset = new HashSet<Integer>(); for (int n : num) { hashset.add(n); } int lcs = 0; for (int n : num) { int i = n, count = 1; hashset.remove(n); // i-- while (hashset.contains(--i)) { count++; hashset.remove(i); } // i++ i = n; while (hashset.contains(++i)) { count++; hashset.remove(i); } // update lcs lcs = Math.max(lcs, count); } return lcs; } } ~~~ ### 源码分析 首先使用 HashSet 建哈希表，然后遍历数组，依次往左往右搜相邻数，搜到了就从 Set 中删除。末尾更新最大值。 ### 复杂度分析 时间复杂度和空间复杂度均为 O(n)O(n)O(n).