128. Longest Consecutive Sequence

128. Longest Consecutive Sequence

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

You must write an algorithm that runs in O(n) time.

Example 1:

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

Example 2:

Input: nums = [0,3,7,2,5,8,4,6,0,1]
Output: 9

Constraints:

• 0 <= nums.length <= 105

• -109 <= nums[i] <= 109

My Solutions:

• 方法1：先排序，再traverse每一个数。Time: O(nlgn)

• 方法2：把nums的每个数存在hashset里。在traverse hashset的时候，把每个数的+1和-1都找一遍。每次找到一个数，都把Ta从set里删去。Time: O(n)

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 res = 0;
        
        for (int i : nums) {
            if (set.contains(i)) {
                set.remove(i);
                
                int pre = i - 1;
                int next = i + 1;
                while (set.contains(pre)) {
                    set.remove(pre);
                    pre--;
                }
                while (set.contains(next)) {
                    set.remove(next);
                    next++;
                }
                res = Math.max(res, next - pre - 1);
            }
            
        }
        return res;
        
    }
}