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 considered overlapping.
1 /** 2 * Definition for an interval. 3 * struct Interval { 4 * int start; 5 * int end; 6 * Interval() : start(0), end(0) {} 7 * Interval(int s, int e) : start(s), end(e) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 static bool sfun(Interval a,Interval b){ //需要用static 13 //2. sort中的比较函数compare要声明为静态成员函数或全局函数，不能作为普通成员函数，否则会报错。 invalid use of non-static member function 14 //因为：非静态成员函数是依赖于具体对象的，而std::sort这类函数是全局的，因此无法再sort中调用非静态成员函数。静态成员函数或者全局函数是不依赖于具体对象的, 可以独立访问，无须创建任何对象实例就可以访问。同时静态成员函数不可以调用类的非静态成员。 15 16 return a.start<b.start; 17 } 18 vector<Interval> merge(vector<Interval>& intervals) { 19 vector<Interval> res; 20 if(intervals.empty()) return res; 21 22 sort(intervals.begin(),intervals.end(),sfun); 23 24 res.push_back(intervals[0]); 25 26 for(int i = 1;i<intervals.size();i++) 27 { 28 Interval pre = res.back();//// 29 Interval now = intervals[i]; 30 if(now.start>pre.end) 31 res.push_back(now); 32 else 33 { 34 Interval ii ; 35 ii.start = pre.start; 36 ii.end = max(now.end,pre.end);// 注意【1,4】【2，3】 这种需要返回【1,4】所以用max 37 res.back() = ii; 38 } 39 } 40 return res; 41 } 42 43 };