Given a collection of intervals, merge all overlapping intervals.
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] overlaps, 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.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
Constraints:
intervals[i][0] <= intervals[i][1]
Solution
class Solution {
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
if (intervals.empty()) return intervals;
auto comparator = [](const auto &a, const auto &b) {
if (a[0] != b[0]) return a[0] < b[0];
else return a[1] < b[1];
};
sort(intervals.begin(), intervals.end(), comparator);
vector<vector<int>> merged; merged.reserve(intervals.size());
merged.push_back(intervals[0]);
for (int i = 1; i < intervals.size(); ++i) {
const int end = merged.back()[1];
const int start = intervals[i][0];
if (start <= end) {
merged.back()[1] = max(end, intervals[i][1]);
} else {
merged.push_back(intervals[i]);
}
}
return merged;
}
};
