题目
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].
分析
给定几个区间,要求合并重叠区间,返回结果;
AC代码
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
//自定义Interval类型元素的升序比较函数
bool cmp(Interval a, Interval b)
{
return a.start < b.start;
}
class Solution {
public:
vector<Interval> merge(vector<Interval>& intervals) {
//如果输入参数为空,则返回空vector
if (intervals.empty())
return vector<Interval>();
int len = intervals.size();
//首先,按照每个Integerval的区间首值进行排序,自定义比较
sort(intervals.begin(), intervals.end() , cmp);
//声明结果
vector<Interval> ret;
vector<Interval>::iterator iter = intervals.begin();
//定义临时变量
Interval temp = intervals[0];
for (int i = 0; i < len; i++)
{
//换一种判断方法
if (intervals[i].start > temp.end)
{
ret.push_back(temp);
temp = intervals[i];
}
else{
temp.end = temp.end > intervals[i].end ? temp.end : intervals[i].end;
}//else
}//for
ret.push_back(temp);
return ret;
}
};