java-Longest Consecutive Sequence

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.

乱序数组中找出最长连续序列 只能用O（N）时间复杂度 ，先排序再查找需要O（NlogN）显然不满足 线性时间查找只有用 哈希表 。再建立一个栈 每次在栈空时从数组中依序赋值com=num[index] 哈希表中删除此项 再用哈希查找 com-1和com+1若有则计数+1 哈希表中删除此项 并且将其加入队列 继续查找 代码如下：

public class Solution {
    public int longestConsecutive(int[] num) {
        if (num.length <= 1)
			return num.length;
		Map<Integer, Integer> hm = new HashMap<Integer, Integer>();
		for (int i = 0; i < num.length; i++) {
			if (!hm.containsKey(num[i])) {
				hm.put(num[i], 1);
			}
		}
		Queue<Integer> que = new LinkedList<Integer>();
		que.offer(num[0]);
		hm.remove(num[0]);
		int index = 1;
		int max = 1;
		int count = 1;
		while (index <=num.length - 1) {
			if (que.isEmpty()) {
				if (hm.containsKey(num[index])) {
					que.offer(num[index]);
					hm.remove(num[index]);
					index++;
				} else {
					index++;
				}
				if(count!=1){
				   max=Math.max(max, count);
				   count=1;
				}
			} else {
				int com = que.poll();
				if (hm.containsKey(com - 1)) {
					hm.remove(com - 1);
					que.offer(com - 1);
					count++;
				}
				if (hm.containsKey(com + 1)) {
					hm.remove(com + 1);
					que.offer(com + 1);
					count++;
				}
			}
		}
		return max;
    }
}