Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
在原始数组上操作，先按照start值在原数组中二分查找待插入的区间，假设查找到的位置为ite，从ite或者ite-1开始合并区间直到不能合并为止（终止条件是合并后区间的end<当前区间的start）,然后在原数组中删除参与合并的区间，再插入合并后的新区
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class
Solution {
private
:
static
bool
comp(Interval a, Interval b)
{
return
a.start < b.start;
}
public
:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
//在原始数组上操作
vector<Interval>::iterator ite = lower_bound(intervals.begin(),intervals.end(), newInterval, comp);
//按照start值二分查找
if
(ite != intervals.begin() && newInterval.start <= (ite-1)->end)
//ite的上一个区间也可能参与合并
{
ite--;
//合并后新区间的起点只和第一个合并的区间有关，因为数组时按区间起点有序的
newInterval.start = min(newInterval.start, ite->start);
}
vector<Interval>::iterator eraseBegin = ite;
for
(; ite != intervals.end() && newInterval.end >= ite->start; ite++)
if
(newInterval.end < ite->end)newInterval.end = ite->end;
//合并后的新区间存放于newInterval
ite = intervals.erase(eraseBegin, ite);
//[eraseBegin, ite)是合并时应该删掉的区间
intervals.insert(ite, newInterval);
//插入合并后的区间
return
intervals;
}
};
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新建数组存放结果
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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> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval> res;
res.reserve(intervals.size());
int
i;
//插入前部分不需要合并的区间
for
(i = 0; i < intervals.size() && intervals[i].end < newInterval.start; i++)
res.push_back(intervals[i]);
//i为需要合并的起点，注意的是合并后新区间的起点只和第一个合并的区间有关，因为数组时按区间起点有序的
if
(i < intervals.size())newInterval.start = min(newInterval.start, intervals[i].start);
//合并区间
for
(; i < intervals.size() && newInterval.end >= intervals[i].start; i++)
if
(newInterval.end < intervals[i].end)newInterval.end = intervals[i].end;
//插入合并后的区间
res.push_back(newInterval);
//插入剩余的区间
res.insert(res.end(), intervals.begin()+i, intervals.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/3715013.html，如需转载请自行联系原作者
