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.

思想： 用Set（HashSet）来确定每一个数相邻的序列。

java代码：

1. public int longestConsecutive(int[] num) {
3. Set<Integer> data = new HashSet<Integer>();
4. int nlen = num.length;
5. if(nlen == 0) return 0;
6. for(int i=0;i<nlen;i++) {
7. data.add(num[i]);
8. }
9. int longest = 0;
10. for(int i=0;i<nlen;i++) {
11. int step = 0;
12. int start = num[i];
13. while(data.contains(start)){
14. data.remove(start);
15. start++;
16. step++;
17. }
18. start = num[i]; //还是上一个起点
19. while(data.contains(start-1)){
20. step++;
21. start--;
22. data.remove(start);
23. }
24. if(step>longest) {
25. longest = step;
26. }
27. }
28. return longest;
29. }

C++代码：了解 unordered_set 和java中的HashMap一致

1. int longestConsecutive(vector<int> &num) {
2. if(num.empty()) return 0;
3. int i=0;
4. int sz = num.size();
5. unordered_set<int> numset;
6. for(i=0;i<sz;i++) {
7. if(numset.count(num[i])==0)
8. numset.insert(num[i]);
9. }
10. int maxlg = 0;
11. for(i=0;i<sz;i++) {
12. int lg = 1;
13. if(numset.count(num[i])!=0) {
14. int c=num[i]+1;
15. while(numset.count(c)!=0) {
16. lg++;
17. numset.erase(c);
18. c++;
19. }
20. c = num[i]-1;
21. while(numset.count(c)!=0) {
22. lg++;
23. numset.erase(c);
24. c--;
25. }
26. }
27. if(lg>maxlg) maxlg = lg;
28. }
29. return maxlg;
30. }