# 56. Merge Intervals
Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
# Solution
Approach 1: Sorting.
Approach 2: Connected components.
# Code (Python)
Approach 1:
class Solution:
def merge(self, intervals: List[List[int]]) -> List[List[int]]:
intervals.sort()
result = []
for interval in intervals:
if result and interval[0] <= result[-1][1]:
result[-1][1] = max(interval[1], result[-1][1])
else:
result.append(interval)
return result
Approach 2:
# Code (C++)
Approach 1:
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
private:
static bool intervalCmp(Interval a, Interval b) {
return a.start < b.start;
}
public:
vector<Interval> merge(vector<Interval>& intervals) {
vector<Interval> merged;
if (intervals.size() == 0)
return merged;
std::sort(intervals.begin(), intervals.end(), intervalCmp);
int head = intervals[0].start;
int tail = intervals[0].end;
for (int i = 1; i < intervals.size(); ++i)
{
if (tail < intervals[i].start)
{
merged.push_back(Interval(head, tail));
head = intervals[i].start;
tail = intervals[i].end;
}
else
{
tail = std::max(tail, intervals[i].end);
}
}
merged.push_back(Interval(head, tail));
return merged;
}
};
Approach 2: