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 };

转载于:https://www.cnblogs.com/TenosDoIt/p/3422249.html

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