128. Longest Consecutive Sequence

最新推荐文章于 2025-09-11 19:30:36 发布

转载最新推荐文章于 2025-09-11 19:30:36 发布 · 62 阅读

文章标签：

#数据结构与算法 #java

本文介绍了一种在未排序整数数组中寻找最长连续元素序列的算法，通过使用HashSet进行高效查找并删除操作，实现时间复杂度为O(n)。文章提供了详细的Java代码实现，并附带了二次复刷的代码优化版本。

题目：

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,
Given [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.

Your algorithm should run in O(n) complexity.

链接： http://leetcode.com/problems/longest-consecutive-sequence/

题解：

把数字加入hashset里，然后向左右查找，同时更新max。

Time Complexity - O(n)， Space Complexity - O(n)

public class Solution {
    public int longestConsecutive(int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;
        HashSet<Integer> set = new HashSet<>();
        for(int i : nums)
            set.add(i);
        int max = 0;
        
        for(int i = 0; i < nums.length; i++) {
            if(set.contains(nums[i])) {
                set.remove(nums[i]);
                int left = nums[i] - 1, right = nums[i] + 1, localMax = 1;
                
                while(set.contains(left)) {
                    set.remove(left);
                    left--;
                    localMax++;
                }
                while(set.contains(right)) {
                    set.remove(right);
                    right++;
                    localMax++;
                }
             max = Math.max(max, localMax);   
            }
        }
            
        return max;
    }
}

二刷:

Java:

Time Complexity - O(n)， Space Complexity - O(n)

public class Solution {
    public int longestConsecutive(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        Set<Integer> set = new HashSet<>();
        for (int num : nums) set.add(num);
        int max = 0;
        for (int num : nums) {
            if (set.contains(num)) {
                int count = 1;
                set.remove(num);
                int i = 1;
                while (set.contains(num - i)) {
                    set.remove(num - i);
                    i++;
                    count++;
                }
                i = 1;
                while (set.contains(num + i)) {
                    set.remove(num + i);
                    i++;
                    count++;
                }
                max = Math.max(max, count);
            }
        }
        return max;
    }
}