# 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: