对所有区间按起点位置进行排序
遍历到k点后的[l,r]:已知的包含k点的最大区间
遍历到k+1点,若该起点属于[l,r]那么可能扩展这个区间的终点,r=max(r,k+1点的end)
/**
* 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) {
vector<Interval> ans;
sort(intervals.begin(), intervals.end(),
[](const Interval a, const Interval b) {return a.start < b.start; });
if (intervals.empty()) return ans;
int l = intervals[0].start, r = intervals[0].end;
for (auto &x : intervals)
if (x.start >= l && x.start <= r) r = max(x.end, r);
else ans.push_back(Interval(l, r)), l = x.start, r = x.end;
ans.push_back(Interval(l, r));
return ans;
}
};