题目链接

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].

对若干个区间进行合并，使合并后的区间没有重叠

先对区间按照左边界排序，然后顺序扫描，合并重叠的区间即可。              本文地址

代码1在原区间数组上操作，不使用额外的空间，但是需要删除多余的区间，这样会移动数组元素

/*** 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 {
private:static bool comp(Interval a, Interval b){return a.start < b.start;}
public:vector<Interval> merge(vector<Interval> &intervals) {if(intervals.empty())return intervals;sort(intervals.begin(), intervals.end(), comp);vector<Interval>::iterator it1 = intervals.begin(), it2 = it1 + 1;while(it1 != intervals.end() && it2 != intervals.end()){if(it2->start <= it1->end){if(it1->end < it2->end)it1->end = it2->end;it2++;}else{//[it1+1, it2)范围内的区间可以从原数组删除it1 = intervals.erase(it1+1, it2);it2 = it1 + 1;}}if(it1 != intervals.end())intervals.erase(it1 + 1, it2);return intervals;}
};

代码2用新数组来存储合并后的区间

/*** 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 {
private:static bool comp(Interval a, Interval b){return a.start < b.start;}
public:vector<Interval> merge(vector<Interval> &intervals) {if(intervals.empty())return intervals;sort(intervals.begin(), intervals.end(), comp);vector<Interval> res;res.push_back(intervals[0]);for(int i = 1; i < intervals.size(); i++){Interval &p = res.back();if(intervals[i].start > p.end)res.push_back(intervals[i]);else if(intervals[i].end > p.end)p.end = intervals[i].end;}return res;}
};

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