题目描述：点击

1 /** 2  * Definition for an interval. 3  * struct Interval { 4  *     int start; 5  *     int end; 6  *     Interval() : start(0), end(0) {} 7  *     Interval(int s, int e) : start(s), end(e) {} 8  * }; 9  */10 class Solution {11 public:12   vector
     
       insert(vector
      
        &intervals, Interval newInterval) {13     // Start typing your C/C++ solution below14     // DO NOT write int main() function15     int maxN = 1000000000;16     vector
       
         rlt;17     if (intervals.empty()){18       rlt.push_back(newInterval);19       return rlt;20     }21     intervals.push_back(newInterval);22     vector
        
         ::iterator vi;23     vector
         
          ::iterator pos =intervals.end()-1;24     for (vi=intervals.end()-1-1; vi+1!=intervals.begin(); vi--){25       if ((*vi).start>(*(vi+1)).start){26         pos = vi;27         Interval tmp = * vi;28         *vi = *(vi+1);29         *(vi+1) = tmp;30       }31     }32     for (vi = pos-1; vi+1!=intervals.begin(); vi--){33       if ( (*vi).end>=(*pos).start ){34         (*pos).start = (*vi).start;35         (*pos).end = (*pos).end > (*vi).end ? (*pos).end : (*vi).end;36         (*vi).start = maxN;37       }38       else break;39     }40     for (vi = pos+1; vi!=intervals.end(); vi++){41       if ( (*vi).start<=(*pos).end ){42         (*pos).end = (*pos).end > (*vi).end ? (*pos).end : (*vi).end;43         (*vi).start = maxN;44       }45       else break;46     }47     for (vi = intervals.begin(); vi!=intervals.end(); vi++){48       if ((*vi).start != maxN)49         rlt.push_back(*vi);50     }51     return rlt;52   }53 };