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].
法一：直接复用之前写的insert函数，依次将区间段插入到结果集中
1 #include "stdafx.h" 2 #include <vector> 3 #include <iostream> 4 #include <algorithm> 5 using namespace std; 6 7 struct Interval 8 { 9 int start; 10 int end; 11 Interval():start(0),end(0) {} 12 Interval(int s, int e):start(s),end(e) {} 13 }; 14 class Solution 15 { 16 public: 17 vector<Interval> merge(vector<Interval> &intervals) 18 {//未排序的区间段数组，进行合并 19 //法一：复用之前的insert函数，每次从intervals中取一个区间插入到结果集中 20 vector<Interval> res; 21 for(int i=0; i<intervals.size(); i++) 22 res = insert(res,intervals[i]); 23 return res; 24 25 } 26 vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) 27 {//参考：书 28 //从前向后比较，看是否插入。前提：区间段已排序,无重合 29 //改进之处：方法没变，但由于insert和erase函数代价有点高，会移动修改，故，不做原地的，空间换时间，直接新建一个好了。 30 vector<Interval> res; 31 int count = intervals.size(); 32 if(count == 0) 33 { 34 res.push_back(newInterval);//防止待插入区间在最后 35 return res; 36 } 37 38 int index = 0; 39 while(index<count) 40 { 41 if(newInterval.end < intervals[index].start) 42 {//当前区间在待插入区间之前，直接插入待插入区间 43 res.push_back(newInterval); 44 while(index<count) 45 { 46 res.push_back(intervals[index]);//剩余元素插入res 47 index++; 48 } 49 return res; 50 } 51 else if(newInterval.start > intervals[index].end) 52 {//当前区间大于待插入区间，跳过，继续判断 53 res.push_back(intervals[index]); 54 } 55 else 56 {//当前区间与待插入区间之间有重合部分 57 newInterval.start = min(newInterval.start,intervals[index].start); 58 newInterval.end = max(newInterval.end,intervals[index].end); 59 } 60 index++; 61 } 62 res.push_back(newInterval);//防止待插入区间在最后 63 return res; 64 } 65 };
法二：将数组自定义排序，然后从前向后两两合并
1 #include "stdafx.h" 2 #include <vector> 3 #include <iostream> 4 #include <algorithm> 5 using namespace std; 6 7 struct Interval 8 { 9 int start; 10 int end; 11 Interval():start(0),end(0) {} 12 Interval(int s, int e):start(s),end(e) {} 13 }; 14 class Solution 15 { 16 public: 17 //自定义的比较函数，为使sort调用，必须是全局或者是静态的，不能是普通成员函数 18 static bool compare(Interval v1, Interval v2) 19 { 20 return v1.start==v2.start ? v1.end<v2.end : v1.start<v2.start; 21 } 22 vector<Interval> merge(vector<Interval> &intervals) 23 {//未排序的区间段数组，进行合并 24 //法二：先将区间段数组排序，然后从前向后两两合并 25 //参考：http://www.cnblogs.com/ganganloveu/p/4158759.html 26 27 vector<Interval> res;//结果集 28 int count = intervals.size(); 29 if(count == 0) 30 return res; 31 //区间段排序 32 sort(intervals.begin(),intervals.end(),compare); 33 res.push_back(intervals[0]); 34 35 for(int i=1; i<count; i++) 36 {//依次插入当前区间段，将当前区间段与结果集res的最后区间段进行比较，看是否合并——直接修改数据即可 37 Interval &back = res.back();//back要是引用，否则不会改变其值!!! 38 if(intervals[i].start > back.end)//无重合 39 res.push_back(intervals[i]); 40 else 41 //有重合 42 back.end = max(back.end, intervals[i].end); 43 } 44 return res; 45 } 46 }; 47 int main() 48 { 49 Solution sol; 50 51 Interval data1[] = {Interval(1,3),Interval(2,6),Interval(8,10),Interval(15,18)}; 52 vector<Interval> test1(data1,data1+4); 53 //test1 54 for(auto &i : test1) 55 cout << "["<<i.start << ","<< i.end<<"]"; 56 cout << endl; 57 vector<Interval> res1 = sol.merge(test1); 58 for(auto &i : res1) 59 cout << "["<<i.start << ","<< i.end<<"]"; 60 cout << endl; 61 cout << endl; 62 63 Interval data2[] = {Interval(3,8),Interval(2,9),Interval(6,7),Interval(8,10),Interval(1,2)}; 64 vector<Interval> test2(data2,data2+5); 65 //test2 66 for(auto &i : test2) 67 cout << "["<<i.start << ","<< i.end<<"]"; 68 cout << endl; 69 vector<Interval> res2 = sol.merge(test2); 70 for(auto &i : res2) 71 cout << "["<<i.start << ","<< i.end<<"]"; 72 cout << endl; 73 74 return 0; 75 }
参考：http://www.cnblogs.com/ganganloveu/p/4158759.html