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：
分3种情况讨论：
首先，如果插入的元素的start一直大于上一个的结束end，那么说明还没找到要插入的位置；
如果找到要插入的位置，即有start小于end。如果要插入的元素的end也小于找到要插入点的start，说明没有交集，直接将元素插入，设置标志位用来控制该新插入的元素只插入一次。
否则，如果插入的元素与后一个区间有交集，则将新插入的元素与后一个区间合并成新的要插入的元素，继续重复上面的过程。记得要判断flag看要插入的元素是否插入了。
C++实现代码：
#include<iostream> #include<vector> #include<algorithm> using namespace std; 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> ret; if(intervals.empty()) { ret.push_back(newInterval); return ret; } int flag=0; int i; for(i=0; i<(int)intervals.size(); i++) { if(newInterval.start>intervals[i].end) { ret.push_back(intervals[i]); } else if(newInterval.end<intervals[i].start) { if(flag==1) ret.push_back(intervals[i]); else { flag=1; ret.push_back(newInterval); ret.push_back(intervals[i]); } } else { newInterval.start=min(newInterval.start,intervals[i].start); newInterval.end=max(newInterval.end,intervals[i].end); } } if(flag==0) ret.push_back(newInterval); return ret; } }; int main() { Solution s; Interval a1(1,2); Interval a2(3,5); Interval a3(6,7); Interval a4(8,10); Interval a5(12,16); vector<Interval> intervals= {a1,a2,a3,a4,a5}; Interval newInternal= {0,0}; vector<Interval> result=s.insert(intervals,newInternal); for(auto a:result) cout<<"[ "<<a.start<<" , "<<a.end<<" ]"<<endl; }
运行结果：
class Solution { public: vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { if(intervals.empty()) return vector<Interval>({newInterval}); vector<Interval> res; int i; int flag=false; for(i=0; i<intervals.size(); ++i) { if(intervals[i].end<newInterval.start) { res.push_back(intervals[i]); } else if(newInterval.end<intervals[i].start) { flag=true; res.push_back(newInterval); break; } else { newInterval.start=min(newInterval.start,intervals[i].start); newInterval.end=max(newInterval.end,intervals[i].end); } } for(;i<intervals.size();i++) res.push_back(intervals[i]); if(!flag) res.push_back(newInterval); return res; } };
