# C++ O(N) solution, easily understand.

• Go through the list, for each integer, we generate a range, this range is represented by a `pair<int,int>` , the first element is the smallest number on the left side, the second element is the largest number on the right side, then stores this range in a map, which is from an integer to its range. We could use this range to calculate the number of a consecutive sequence.

`````` int longestConsecutive(vector<int>& nums) {
// this map is from an integer to its range.
unordered_map<int,pair<int, int>> range ;
// record the longest consecutive sequence
int maxCount = 0;
for(auto n : nums)
{
if(range.find(n) == range.end())// if the integer we have meet , we don't need to recalculate its range
{
pair<int,int> tmp = {n,n}; // initial its range with itself
range[n] = tmp;
if(range.find(n-1) != range.end())  // if n - 1 existed, update it's smallest number with the smallest number in the range of n-1
{
tmp.first = range[n-1].first;
}
if(range.find(n+1) != range.end())// if n + 1 existed, updates it's largest number with the largest number in the range of n + 1
{
tmp.second = range[n+1].second;
}
range[tmp.first].second = tmp.second; //update the largest number in the range of smallest number with current largest number
range[tmp.second].first = tmp.first; //update the smallest number in the range of largest number with current smallest number
maxCount = max(maxCount, tmp.second - tmp.first + 1);
}
}
return maxCount;

}
``````

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