Longest Consecutive Sequence

题目描述

2.1.6 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

题目分析

解题思路

无序数列中最长连续的数 长度，时间复杂度要求为O(n)
1、先排序 时间复杂度为 O(nlogn) 这个时间复杂度是怎么的出来的
2、使用哈希表 unordered_map<int, bool> used 记录每个元素是否使用，对没元素扩散查找，直到不连续记录下长度

代码实现

/*
date:2018/3/06 16:59
*/

#include <vector>
#include <unordered_map>
#include <algorithm>
using std::unordered_map;
using std::vector;
using std::max;
int arr[] = {1,2,3,8,4,10,6,7};
int arr_len = sizeof(arr)/sizeof(int);

vector<int> nums(arr,arr+arr_len);

int longestConsecutive(vector<int>&nums)
{
int len = 0;
int result = 0;
//初始化哈希表
unordered_map<int,bool> used;
for(int i=0; i<nums.size(); i++)
used[nums[i]] = false;
//进行遍历
for (int i=0; i<nums.size(); i++)
{
int value = nums[i];
unordered_map<int,bool>::iterator it = used.find(value);
while (it!=used.end() && !it->second )
{
len++;
value++;
it->second = true;
it = used.find(value);
}
printf("begin [%d] to",len ? value-1 : value);
value = nums[i]-1;
it = used.find(value);
while (it!=used.end() && !it->second )
{
len++;
value--;
it->second = true;
it = used.find(value);
}
printf("end [%d],len[%d]\n",value+1,len);
result = max(len,result);
len = 0;
}
return result;
}

参考

https://www.geek-share.com/detail/2670132920.html //map hashmap unordered_map
https://www.geek-share.com/detail/2681020403.html
http://www.cplusplus.com/reference/unordered_map/unordered_map/find/