LeetCode(128)Longest Consecutive Sequence

题目如下：

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.

分析如下：

因为要求了O(n)的复杂度，所以不能排序之后来做, 所以要考虑空间换时间，所以考虑hash。

我的代码：

class Solution {
public:
    int longestConsecutive(vector<int> &num) {
        std::unordered_map<int, int> map1;
        std::unordered_map<int, int>::iterator it1;
        int cur_key = 0;
        int max_length = 1;
        int cur_length = 1;
        for (int i = 0; i < num.size(); ++i) {
            map1[num[i]]=1;
        }
        
        //for (it1 = map1.begin(); it1 != map1.end(); ++it1) {//for循环中进行删除会使得迭代器失效 demo
        while (map1.size()>1) {
            it1 = map1.begin();  // 以it1指向的元素(it1->first)作为基准点，分别向左边和右边进行扩展。
            cur_key = it1->first+1;// 开始右边的搜索
            while (map1.find(cur_key) != map1.end()) {
                map1.erase(cur_key); //删除，提高后续搜索的效率
                cur_key++;
            }
            cur_length = cur_key - it1->first; // 右侧长度

            cur_key = it1->first - 1; //开始左边的搜索
            while (map1.find(cur_key) != map1.end()) {
                map1.erase(cur_key);
                cur_key--;
            }
            cur_length = cur_length + it1->first - cur_key - 1;//左侧+右侧的长度
            if (cur_length > max_length)
                max_length = cur_length;
            map1.erase(it1->first); //NOTE: 不要忘记删除基准点，否则当有很多个离散的基准点时会infinite loop.
        }
        return max_length;
    }
};