[leetcode] 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.

这题我的第一反应是bitmap，但mapping过去之后，怎么搜这个bitmap又是一个很大的问题。

其实用hash就可以了，首先把数组放在hash表里， O(n)

从表里拿到一个元素之后查找它的前后元素在不在hash表中，找到了把当前元素删除。

每次查找和删除都是O(1)，所以hash表的遍历复杂度也是O(n)

顺便贴一下STL里面hash删除的用法

by position (1)

iterator erase ( const_iterator position );

by key (2)

size_type erase ( const key_type& k );

range (3)

iterator erase ( const_iterator first, const_iterator last );

Versions (1)  (3) 返回被删除元素的下一个元素的iterator
Version (2) 返回被删除的元素个数

class Solution {public:    int longestConsecutive(vector<int> &num) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        unordered_set<int> hash;                for(int i = 0; i < num.size(); i++){            hash.insert(num[i]);        }                int ret = 0;        int cnt;                unordered_set<int>::iterator it;        int cur_elem;        int tmp_elem;                while(!hash.empty()){            it = hash.begin();            cur_elem = *it;            hash.erase(it);                        tmp_elem = cur_elem;            cnt = 1;                        while((it = hash.find(tmp_elem + 1))!= hash.end()){                hash.erase(it);                tmp_elem++;                cnt++;            }                        tmp_elem = cur_elem;            while((it = hash.find(tmp_elem - 1))!= hash.end()){                hash.erase(it);                tmp_elem--;                cnt++;            }                        ret = max(ret, cnt);        }                return ret;    }};