题目：
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.
题解：
这道题利用HashSet的唯一性解决，能使时间复杂度达到O(n)。首先先把所有num值放入HashSet，然后遍历整个数组，如果HashSet中存在该值，就先向下找到边界，找的同时把找到的值一个一个从set中删去，然后再向上找边界，同样要把找到的值都从set中删掉。所以每个元素最多会被遍历两边，时间复杂度为O(n)。
代码如下：
1 public int longestConsecutive(int[] num) {
2 if(num == null||num.length == 0)
3 return 0;
4
5 HashSet<Integer> hs = new HashSet<Integer>();
6
7 for (int i = 0 ;i<num.length; i++)
8 hs.add(num[i]);
9
10 int max = 0;
11 for(int i=0; i<num.length; i++){
12 if(hs.contains(num[i])){
13 int count = 1;
14 hs.remove(num[i]);
15
16 int low = num[i] - 1;
17 while(hs.contains(low)){
18 hs.remove(low);
19 low--;
20 count++;
21 }
22
23 int high = num[i] + 1;
24 while(hs.contains(high)){
25 hs.remove(high);
26 high++;
27 count++;
28 }
29 max = Math.max(max, count);
30 }
31 }
32 return max;
33 }
2 if(num == null||num.length == 0)
3 return 0;
4
5 HashSet<Integer> hs = new HashSet<Integer>();
6
7 for (int i = 0 ;i<num.length; i++)
8 hs.add(num[i]);
9
10 int max = 0;
11 for(int i=0; i<num.length; i++){
12 if(hs.contains(num[i])){
13 int count = 1;
14 hs.remove(num[i]);
15
16 int low = num[i] - 1;
17 while(hs.contains(low)){
18 hs.remove(low);
19 low--;
20 count++;
21 }
22
23 int high = num[i] + 1;
24 while(hs.contains(high)){
25 hs.remove(high);
26 high++;
27 count++;
28 }
29 max = Math.max(max, count);
30 }
31 }
32 return max;
33 }