LeetCode | 57. Insert Interval

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].

题目大意：给定一些区间（已经按start升序排列），和一个要你插入的区间，插入这个目标区间，并做相应的合并。

麻烦之处：要注意的细节比较多。比如将区间插入空的区间集合；插入之后做完合并之后还要保持start值有序等。总体难度评级为medium吧，应该算不上hard。

// 16 ms
/**
* 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(Interval a,Interval b)
{
return a.start<b.start;
}

vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
int flag = 0;       //判断该区间是否放入
int n = intervals.size();
vector<Interval> res;

if(n == 0)
{
res.push_back(newInterval);
return res;
}
if(newInterval.end < intervals[0].start)        //放在最左边，因为答案要求 x 坐标有序。。。
{
flag = 1;
res.push_back(newInterval);
}

for(int i=0;i<n;i++)
{
if(intervals[i].end < newInterval.start || intervals[i].start>newInterval.end)      //不会重合
res.push_back(intervals[i]);
else if(flag == 0)
{
flag = 1;
Interval tmp = intervals[i];
tmp.start = min(tmp.start,newInterval.start);       //更新左端点
while(i<n && intervals[i].start<=newInterval.end)   //有重合
{
tmp.end = max(intervals[i].end,newInterval.end);
i++;
}
res.push_back(tmp);
i--;
}
}
if(flag == 0)
res.push_back(newInterval);

sort(res.begin(),res.end(),cmp);        //答案要有序

return res;
}
};