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
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/**
* 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;
}
};
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代码2用新数组来存储合并后的区间
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/**
* 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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本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3714681.html，如需转载请自行联系原作者
