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：

class Solution {
public:
    int longestConsecutive(vector<int> &num) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int res = 1, len = num.size();
        if(len == 0)return 0;
        sort(num.begin(), num.end());
        int curr = 1;
        for(int i = 1; i < len; i++)
        {
            if(num[i] - num[i-1] == 1)
            {
                curr++;
                if(curr > res)res = curr;
            }
            else if(num[i] - num[i-1] == 0);
            else 
                curr = 1;
        }
        return res;
    }
};

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

class Solution {
public:
    int longestConsecutive(vector<int> &num) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int res = 1, len = num.size();
        if(len == 0)return 0;
        unordered_set<int> hashtable;
        for(int i = 0; i < len; i++)
            hashtable.insert(num[i]);
        while(hashtable.empty() == false)
        {
            int currlen = 1;
            int curr = *(hashtable.begin());
            hashtable.erase(curr);
            int tmp = curr-1;
            while(hashtable.empty()==false && 
                hashtable.find(tmp) != hashtable.end())
            {
                hashtable.erase(tmp);
                currlen++;
                tmp--;
            }
            tmp = curr+1;
            while(hashtable.empty()==false && 
                hashtable.find(tmp) != hashtable.end())
            {
                hashtable.erase(tmp);
                currlen++;
                tmp++;
            }
            if(res < currlen)res = currlen;
        }
        return res;
    }
};

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