一天一道LeetCode系列

（一）题目

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].

（二）解题

相关题目：

/*主要解题思路：1、当intervals[i].end
   
    newInterval.start的时候，确定合并后的interval的start值3、当找到最后一个满足intervals[i].start>newInterval.end的时候，确定合并后的interval的end值*//** * 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
    
      insert(vector
     
      & intervals, Interval newInterval) {        vector
      
        ret;        if(intervals.size()==0){
  //特殊情况            ret.push_back(newInterval);            return ret;        }        int i = 0;        Interval tmp;        while(i
       
        
         newInterval.start的i值  //这里需要注意当newInterval比vector里面的值都大的情况 tmp.start = min(i==intervals.size()?newInterval.start:intervals[i].start,newInterval.start); tmp.end = newInterval.end; while(i
         
          <=newInterval.end) { tmp.end = max(intervals[i].end,newInterval.end);//找到合并后的end值 i++; } ret.push_back(tmp); while(i