128. Longest Consecutive Sequence

最新推荐文章于 2024-06-13 05:26:23 发布

BigFatSheep

最新推荐文章于 2024-06-13 05:26:23 发布

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分类专栏： Leetcode Hard

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本文介绍了一种在未排序整数数组中寻找最长连续元素子序列的算法，该算法的时间复杂度为O(n)，并详细解释了如何使用哈希映射存储数字及其子序列长度，以高效地更新和查找子序列。

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Description

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

Your algorithm should run in O(n) complexity.

Example:

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

Problem URL

Solution

给一个未排序数组，找到最长的连续子集的长度。要求常数时间。

Using a map to store number and its subsequence’s length. For nums not in map, get its left neighbor and right neighbor’s subsequence length. Update it to res and map and its edge, because they are in the same subsequence. We must keep head and tail element have right updated length value.

If this num is in map, just continue. Of course, it is optional.

Code

class Solution {
    public int longestConsecutive(int[] nums) {
        int res = 0;
        //a map from num to its subsequence's length.
        Map<Integer, Integer> map = new HashMap<>();
        for (int i : nums){
            if (!map.containsKey(i)){
                //get neighbors subsequence's length if exist.
                int left = map.containsKey(i - 1) ? map.get(i - 1) : 0;
                int right = map.containsKey(i + 1) ? map.get(i + 1) : 0;
                //calcualte new list length, if no neighobrs, 1;
                int sum = left + right + 1;
                res = Math.max(res, sum);
                //put new list length into map, update other edge's length value
                map.put(i, sum);
                map.put(i - left, sum);
                map.put(i + right, sum);
            }
            else{
                //duplicate circumstance;
                continue;
            } 
        }
        return res;
    }
}