56. Merge Intervals

/*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].思路：     先按照集合中个区间的左边界的大小排序，升序； 若 i<j,i、j为第几个区间，则i.start<=j.start 若 i.end >= j.start 则合并为[i.start,j.end]若 i.end < j.start 则不合并，把区间添加到集合中重复以上步骤*//** * 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:static bool comp(Interval x,Interval y){return x.start < y.start || (x.start==y.start && x.end < y.end);}    vector<Interval> merge(vector<Interval>& intervals) {        vector<Interval> res;int n= intervals.size();if(n==0) return res;sort(intervals.begin(),intervals.end(),Solution::comp);int start = intervals[0].start,end= intervals[0].end;for(int i=1;i<n;i++){if(end >= intervals[i].start)//区间重合end = end < intervals[i].end ? intervals[i].end : end;else{res.push_back(Interval(start,end));//记录区间start=intervals[i].start;end=intervals[i].end;}}res.push_back(Interval(start,end));//记录最后一个区间return res;    }};