LeetCode:Longest Consecutive Sequence

题目链接

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.

分析：

算法1：首先想到的是排序，排序后遍历一遍就可以找出最长连续序列的长度，只是要稍微注意下判断连续序列的过程中有可能两个元素相同，比如1 2 2 3，排序复杂度n*log(n)，虽然题目要求O(n)复杂度，但是这个解法也可以通过OJ，代码如下：

 1 class Solution {

 2 public:

 3     int longestConsecutive(vector<int> &num) {

 4         // IMPORTANT: Please reset any member data you declared, as

 5         // the same Solution instance will be reused for each test case.

 6         int res = 1, len = num.size();

 7         if(len == 0)return 0;

 8         sort(num.begin(), num.end());

 9         int curr = 1;

10         for(int i = 1; i < len; i++)

11         {

12             if(num[i] - num[i-1] == 1)

13             {

14                 curr++;

15                 if(curr > res)res = curr;

16             }

17             else if(num[i] - num[i-1] == 0);

18             else 

19                 curr = 1;

20         }

21         return res;

22     }

23 };

View Code

算法2：想要O(n)的算法，我们只有以时间换空间，先把数组中所有元素映射到哈希表。然后以题目给出的数组为例：对于100，先向下查找99没找到，然后向上查找101也没找到，那么连续长度是1，从哈希表中删除100；然后是4，向下查找找到3,2,1，向上没有找到5，那么连续长度是4，从哈希表中删除4,3,2,1。这样对哈希表中已存在的某个元素向上和向下查找，直到哈希表为空。算法相当于遍历了一遍数组，然后再遍历了一遍哈希表，复杂的为O(n)。代码如下：                            本文地址

 1 class Solution {

 2 public:

 3     int longestConsecutive(vector<int> &num) {

 4         // IMPORTANT: Please reset any member data you declared, as

 5         // the same Solution instance will be reused for each test case.

 6         int res = 1, len = num.size();

 7         if(len == 0)return 0;

 8         unordered_set<int> hashtable;

 9         for(int i = 0; i < len; i++)

10             hashtable.insert(num[i]);

11         while(hashtable.empty() == false)

12         {

13             int currlen = 1;

14             int curr = *(hashtable.begin());

15             hashtable.erase(curr);

16             int tmp = curr-1;

17             while(hashtable.empty()==false && 

18                 hashtable.find(tmp) != hashtable.end())

19             {

20                 hashtable.erase(tmp);

21                 currlen++;

22                 tmp--;

23             }

24             tmp = curr+1;

25             while(hashtable.empty()==false && 

26                 hashtable.find(tmp) != hashtable.end())

27             {

28                 hashtable.erase(tmp);

29                 currlen++;

30                 tmp++;

31             }

32             if(res < currlen)res = currlen;

33         }

34         return res;

35     }

36 };

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