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) {} * }; */ bool sortM(const Interval &a1, const Interval &a2) { return a1.start < a2.start ; } class Solution { public: vector<Interval> merge(vector<Interval> &intervals) { int n = intervals.size(); if(intervals.size() <2) return intervals; sort(intervals.begin(),intervals.end(),sortM); int i,max; Interval next,val = intervals[0]; vector<Interval> result; for(i = 1; i< n; i++) { if(intervals[i].start > val.end){ result.push_back(val); val = intervals[i] ; }else{ val.end = val.end >intervals[i].end ? val.end : intervals[i].end; } } //end of for i result.push_back(val) ; return result ; } };