Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18]
,
return [1,6],[8,10],[15,18]
.
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */ /** 采用自定义的比较函数,必须申明为全局对象,不能是类的局部对象。 */ bool comp(const Interval &a, const Interval &b) { if(a.start==b.start)return a.end<b.end; return a.start<b.start; } class Solution { public: vector<Interval> merge(vector<Interval> &intervals) { vector<Interval> res; if(intervals.empty())return res; sort(intervals.begin(),intervals.end(),comp); Interval temp(intervals[0].start, intervals[0].end); for (int i=1;i<intervals.size();i++) { if (intervals[i].start<=temp.end) { temp.end=temp.end>intervals[i].end?temp.end:intervals[i].end; } else { res.push_back(temp); temp.start=intervals[i].start; temp.end=intervals[i].end; } } res.push_back(temp); return res; } };