【leetcode】Insert Interval

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].

 

 

 /**
  * 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:
     vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {

         int n=intervals.size();
         vector<Interval> result;

         if(n==)
         {
             result.push_back(newInterval);
             return result;
         }

         int start=newInterval.start;
         int end=newInterval.end;

         bool isfindStart=false;
         bool isfindEnd=false;
         bool addCur=true;

         for(int i=;i<n;i++)
         {
             addCur=true;

             if(!isfindStart)
             {
                 //和当前区间没有交叉,且在start的前面
                 if(start>intervals[i].end)
                 {
                     //需要添加当前元素
                     addCur=true;
                     //当前元素为最后一个元素时，则认为start找到了
                     if(i==n-)  isfindStart=true;
                 }
                 else
                 {
                     //和当前区间有交叉 或者start小于当前区间

                     start=min(start,intervals[i].start);
                     //后续就不用再找start了
                     isfindStart=true;

                     //如果start和end没有和其他Interval有交叉
                     if(end<intervals[i].start) addCur=true;
                     else addCur=false;
                 }
             }

             if(!isfindEnd)
             {
                 //需要继续找end
                 if(end>intervals[i].end)
                 {

                     if(start<=intervals[i].end) addCur=false;

                     if(i==n-)
                     {
                         isfindEnd=true;

                         if(addCur) result.push_back(intervals[i]);
                         result.push_back(Interval(start,end));

                         continue;
                     }
                 }
                 else
                 {
                     //end小于当前区间，此时找到了end
                     end=end>=intervals[i].start?intervals[i].end:end;

                     //找到了end
                     isfindEnd=true;
                     if(end>=intervals[i].start) addCur=false;

                     result.push_back(Interval(start,end));
                     if(addCur) result.push_back(intervals[i]);

                     continue;
                 }
             }

             if(addCur) result.push_back(intervals[i]);
         }
         return result;
     }
 };