56. Merge Intervals

Question

Given an array of intervals where intervals[i] = [start i , end i ] , merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input .
Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]] Output: [[1,5]] Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Example 3:

Input: intervals = [[4,7],[1,4]] Output: [[1,7]] Explanation: Intervals [1,4] and [4,7] are considered overlapping.

Constraints:

1 <= intervals.length <= 10 4
intervals[i].length == 2
0 <= start i <= end i <= 10 4

Solutions

vector<vector<int>> merge(vector<vector<int>> & intervals) {
	int n = static_cast<int>(intervals.size());
	// when n == 0 or n == 1, it's not necessary to merge.
	if (n <= 1) return intervals;
	
	sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b) {
		return a[0] < b[0];
	});
	
	vector<vector<int>> result;
	
	for (int i = 1; i < n; ++i) {
		auto& prevVec = intervals[i - 1];
		auto& currVec = intervals[i];
		if (prev[1] >= currVec[0]) {
			// merge last element into current vector
			currVec[0] = min(currVec[0], prevVec[0]);
			currVec[1] = max(currVec[1], prevVec[1]);
		} else {
			result.push_back(prevVec);
		}
	}
	
	// there's one last vector remining, store it to the result
	result.push_back(intervals[n - 1]);
	
	return result;
}