Longest Consecutive Sequence

1.问题描述

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.

2.问题分析

题目是让我们求一个整数数组中，存在最长连续整数序列的长度。同时要是O(n)解法。考虑到这个数组是无序的，可以考虑用unordered_map<int, bool>来作数据结构表示该整数数组中数的遍历状态。true为访问过，false为初始态未访问的。接下来我们只要遍历这个无序map，对于每一个原数组中出现的数i，判断i++/i- -是否也存在，若存在则给最长长度加1即可，同时状态要从false修改为访问过的true。因此，只需要O(n)即可求解。

3.源代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <unordered_map>
using namespace std;

class MySolution
{
public:
    int the_longest_consecutive(const vector<int> &num)
    {
        int longest_length = 0;
        unordered_map<int, bool> used;
        for (auto i : num)
            used[i] = false;

        for (auto i : num)
        {
            if (used[i])
                continue;

            int cur_length = 1;
            used[i] = true;

            for (int j = i + 1; used.find(j) != used.end(); j++)
            {
                cur_length++;
                used[j] = true;
            }   
            for (int j = i - 1; used.find(j) != used.end(); j--)
            {
                cur_length++;
                used[j] = true;
            }

            longest_length = max(longest_length, cur_length);
        }
        return longest_length;
    }

};

int main()
{
    vector<int> num = {100, 4, 200, 1, 3, 2};
    MySolution test;
    cout << test.the_longest_consecutive(num) << endl;
    return 0;
}

4.reference

[1]Creative Commons: http://creativecommons.org/licenses/by-nc-sa/3.0/
[2]https://github.com/soulmachine/leetcode