problem:
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]
.
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题意:和56题类似,只是要插入一个新的区间
thinking:
如果采用multimap<int, int>排序,只需将newInterval插入到multimap中即可,具体参考56题解法
code:
class Solution { public: vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { vector<Interval> ret; multimap<int,int> map_intervals; if(intervals.size()==0) //边界条件有点变化 { ret.push_back(newInterval); return ret; } for(vector<Interval>::iterator it=intervals.begin();it!=intervals.end();it++) map_intervals.insert(make_pair((*it).start,(*it).end)); map_intervals.insert(make_pair(newInterval.start,newInterval.end));//就加了这一行代码 multimap<int, int>::iterator p=map_intervals.begin(); Interval tmp(p->first,p->second); for(multimap<int, int>::iterator k=++p;k!=map_intervals.end();k++) { if(k->first<=tmp.end) tmp.end=max(tmp.end,k->second); else { ret.push_back(tmp); tmp.start=k->first; tmp.end=k->second; } } ret.push_back(tmp); return ret; } };
时间: 2024-11-25 13:34:05