Leetcode: Longest Consecutive Sequence

最新推荐文章于 2019-07-12 13:00:00 发布

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本文介绍了一种寻找整数数组中最长连续元素序列的算法，并提供了两种解决方案：一种为O(n)复杂度，另一种为O(n log n)复杂度。通过实例展示了如何实现这两种算法。

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.

Solution1 O(n)

import java.util.Hashtable;
public class Solution {
    public int longestConsecutive(int[] num) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if(num.length < 1)
			return 0;
		Hashtable<Integer, Integer> hashtable = new Hashtable<Integer, Integer>();
		for(int i = 0; i < num.length; i++){
			if(hashtable.containsKey(num[i]))
				hashtable.put(num[i], hashtable.get(num[i]) + 1);
			else
				hashtable.put(num[i], 1);
		}
		int max = 1;
		for(int i = 0; i < num.length; i++){
			if(hashtable.containsKey(num[i])){
				int temp = 1, cur = num[i];
				hashtable.remove(cur);
				temp += getDis(hashtable, cur + 1, true);
				temp += getDis(hashtable, cur - 1, false);
				max = temp > max ? temp : max;
			}
		}
		return max;
	}
	public int getDis(Hashtable<Integer, Integer> hashtable, int cur, boolean inc){
		int dis = 0;
		while(hashtable.containsKey(cur)){
			hashtable.remove(cur);
			dis++;
			if(inc)
				cur++;
			else
				cur--;
		}
		return dis;
	}
}

Solution2 O(nlgn)

public class Solution {
    public int longestConsecutive2(int[] num) {
		// Start typing your Java solution below
		// DO NOT write main() function
		Arrays.sort(num);
		int longest = 1;
		int cur = 1;
		for(int i = 1; i < num.length; i++){
			if(num[i] == num[i - 1] + 1){
				cur++;
				longest = longest > cur ? longest : cur;
			}
			else if(num[i] == num[i - 1])
				continue;
			else
				cur = 1;
		}
		return longest;
	}
}