插入时间段 Insert Interval
题目：
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].
思路：注意这里说了原时间段集合是不存在重叠的，因此，我只需要判断新添时间段对原时间段集合的影响。
1、只是新添加了一个时间段在其中，不影响原有的时间段；
2、影响了一个时间段；
2、合并了两个或多个时间段；
这样看来，只需要按顺序遍历一次有序的时间段集合即可。
/**
* 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 {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval>::iterator it = intervals.begin();
while(it != intervals.end())
{
if(newInterval.end < it->start)
{
intervals.insert(it, newInterval);
return intervals;
}
else if(newInterval.start > it->end)
{
it++;
}
else //发现重叠
{
newInterval.start = min(newInterval.start, it->start);
newInterval.end = max(newInterval.end, it->end);
it = intervals.erase(it); //删除it所指结点，返回下一个结点位置
}
}
intervals.insert(intervals.end(), newInterval);
return intervals;
}
};上面的代码看似时间复杂度是O(N)，实际上对vector的erase的每一次操作都是O(N)的，因此总的时间复杂度是O（N^2）。为了避免这样，不再原vector上做修改，而是新建一个新的vector作为新的时间段集合。以空间换取时间，这样时间和空间复杂度都为O(N)。
/**
* 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 {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval> result;
vector<Interval>::iterator it = intervals.begin();
while(it != intervals.end())
{
if(newInterval.end < it->start)
{
break;
}
else if(newInterval.start > it->end)
{
result.push_back(*it);
it++;
}
else //it结点和新时间段重叠，it结点被时间段影响
{
newInterval.start = min(newInterval.start, it->start);
newInterval.end = max(newInterval.end, it->end);
it++;
}
}
result.push_back(newInterval);
for(;it!=intervals.end();++it)
result.push_back(*it);
return result;
}
};
/**
* 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 {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
multimap<int, int> newlist;
newlist.insert(make_pair(newInterval.start, 1));
newlist.insert(make_pair(newInterval.end, 0));
vector<Interval>::iterator it = intervals.begin();
for(;it != intervals.end();it++)
{
newlist.insert(make_pair(it->start, 1));
newlist.insert(make_pair(it->end, 0));
}
intervals.clear();
multimap<int, int>::iterator it2 = newlist.begin();
int left, right, count;
count = 0;
left = it2->first;
while(it2 != newlist.end())
{
if(it2->second == 1)
count++;
else
count--;
if(count == 0)
{
right = it2->first;
it2++;
if(it2 != newlist.end() && right == it2->first)
{
it2++;
count++;
continue;
}
Interval tmp = Interval(left, right);
intervals.push_back(tmp);
if(it2 != newlist.end())
left = it2->first;
}
else
{
it2++;
}
}
return intervals;
}
};