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 considerred overlapping.
AC code:
/**
* 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 {
public:
vector<Interval> merge(vector<Interval>& intervals) {
if (intervals.empty()) return {};
vector<Interval> ans;
sort(intervals.begin(), intervals.end(),
[](const Interval& a, Interval& b) {
return a.start < b.start;
});
for (const auto& interval : intervals) {
if (ans.empty() || interval.start > ans.back().end) {
ans.push_back(interval);
} else {
ans.back().end = max(ans.back().end, interval.end);
}
}
return ans;
}
};
原文：https://www.cnblogs.com/ruruozhenhao/p/9818599.html