Longest Consecutive Sequence

最新推荐文章于 2024-08-13 15:34:52 发布

转载最新推荐文章于 2024-08-13 15:34:52 发布 · 269 阅读

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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。通过使用哈希映射或排序加搜索的方法，可以在O(N)或O(NlgN)的时间复杂度内解决问题，并详细展示了两种方法的具体实现。

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.

1. HashMap. 时间: O(N) 空间: O(N)

2. sort + search. 时间: O(NlgN) 空间：O(1) ***注意sort后前后两个元素相等的情况！！

public class Solution {
    public int longestConsecutive(int[] num) {
        //map num[i] to 0/1 (represent that if it has been visited)
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i : num) {
            map.put(i, 0);
        }
        
        int max = 1;
        
        for (int i : num) {
            if (map.get(i) == 1) continue;
            
            int temp = i;
            int thismax = 1;
            
            while (map.containsKey(temp + 1)) {
                thismax++;
                temp++;
                map.put(temp, 1);
            }
            
            temp = i;
            while (map.containsKey(temp - 1)) {
                thismax++;
                temp--;
                map.put(temp, 1);
            }
            
            max = Math.max(max, thismax);
        }
        return max;
    }
}

public class Solution {
    public int longestConsecutive(int[] num) {
       Arrays.sort(num);
       int max = 1;
       int thismax = 1;
       for (int i = 1; i < num.length; i++) {
           if (num[i] == num[i-1]) continue;
           if (num[i]-1 == num[i-1]) {
               thismax++;               
               max = Math.max(max, thismax);
           } else {
               thismax = 1;
           }
       }
       return max;
    }
}