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：
Insert Interval是Merge Intervals的一个延伸问题，先看看怎么Merge
Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].
1 vector<Interval> merge(vector<Interval> &intervals) { 2 if(intervals.size() <= 1) 3 return intervals; 4 5 vector<Interval> vres; 6 sort(intervals.begin(), intervals.end(), intvalcomp);//先对interval排序 7 Interval tmp(intervals[0]); 8 for(Interval it:intervals){ 9 if(tmp.start == it.start){ 10 tmp.end = it.end; 11 }else if(tmp.end >= it.start){//intervals有序，必然有tmp.start < it->start 12 if(tmp.end < it.end)//直接无视后者{[1,4],[2,3]} 13 tmp.end = it.end;//直接吞并后者{[1,3],[2,4]} 14 }else{ 15 vres.push_back(tmp);//不相交 16 tmp = it; 17 } 18 } 19 vres.push_back(tmp);//漏掉这句会fail{[1,4],[1,4]} 20 return vres; 21 } 22 23 bool intvalcomp(Interval a, Interval b){ 24 if(a.start == b.start) 25 return a.end < b.end; 26 else 27 return a.start < b.start; 28 }
现在有了这个Merge好了的不相交区间序列，怎么进行插入呢？Insert Interval条件太多，每一个大小等号比较，每一个小下标就能让人栽跟斗，因此它也是我目前最讨厌的题目，没有之一。
一开始尝试这种思路：
“新序列按照start排好序（start肯定是各不相同的），第一步我们先用二分找出有交集的序列片段的开始，这一点很像Search Insert Position，然后再往后处理。”
脑子不清楚憋了一下午，恶心的我两天不能刷Leetcode，如果真要写出来的话，就老老实实下面这样，效率不一定差，因为看题目反正是不想要你改变输入参数，横竖都得遍历一遍来拷贝。挺有意思的是，晚上我看到了Google Campus的youku视频，讲述的就是一个倒霉孩子花了30min写二分Insert Interval的反例。。。
1 vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { 2 vector<Interval> rs; 3 int i = 0; 4 while(i < intervals.size() && intervals[i].end < newInterval.start){//找到第一个起点 5 rs.push_back(intervals[i++]); 6 } 7 if(i == intervals.size()){//为空或过了结尾点 8 rs.push_back(newInterval); 9 return rs; 10 } 11 12 newInterval.start = min(newInterval.start, intervals[i].start); 13 while(i < intervals.size() && intervals[i].start <= newInterval.end){//找到结束点 14 newInterval.end = max(newInterval.end, intervals[i++].end); 15 } 16 rs.push_back(newInterval); 17 18 while(i < intervals.size()){ 19 rs.push_back(intervals[i++]); 20 } 21 return rs; 22 }