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]
.
插入区间并合并,
有一个偷懒的方法,把新区间加进去,利用上一题的合并重叠区间合并。
然而为了显示强大的基本功,我作死写模拟了,结果就是真的各种没有想到的情况。
/**
* 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:
static bool cmp(const struct Interval &a,const struct Interval &b)
{
if(a.start!=b.start)
return a.start<b.start;
return a.end<b.end;
}
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval)
{
vector<Interval> result;
sort(intervals.begin(),intervals.end(),cmp);
int flag=0; //表示插入区间是否加入
for(int i=0;i<intervals.size();i++)
{
if(flag)
result.push_back(intervals[i]); //已插入
else
{
if(newInterval.end<intervals[i].start) //插入区间在第一个并且和其他区间没有交集
{
flag=1;
result.push_back(newInterval);
result.push_back(intervals[i]);
}
else
{
if(newInterval.start<=intervals[i].end) // 有交集
{
flag=1;
intervals[i].start=min(intervals[i].start,newInterval.start);
intervals[i].end=max(intervals[i].end,newInterval.end);
int jump=1;
while(i+jump<intervals.size() && intervals[i+jump].start<=intervals[i].end)//合并后续区间
{
intervals[i].end=max(intervals[i].end,intervals[i+jump].end);
jump++;
}
result.push_back(intervals[i]);
i+=jump-1;
}
else
result.push_back(intervals[i]);//无交集
}
}
}
if(!flag) result.push_back(newInterval);//未插入区间
return result;
}
};