【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].

 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<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
13        
14         int n=intervals.size();
15         vector<Interval> result;
16        
17         if(n==0)
18         {
19             result.push_back(newInterval);
20             return result;
21         }
22        
23         int start=newInterval.start;
24         int end=newInterval.end;
25  
26        
27         bool isfindStart=false;
28         bool isfindEnd=false;
29         bool addCur=true;
30        
31         for(int i=0;i<n;i++)
32         {
33             addCur=true;
34            
35             if(!isfindStart)
36             {
37                 //和当前区间没有交叉,且在start的前面
38                 if(start>intervals[i].end)
39                 {
40                     //需要添加当前元素
41                     addCur=true;
42                     //当前元素为最后一个元素时，则认为start找到了
43                     if(i==n-1)  isfindStart=true;
44                 }
45                 else
46                 {
47                     //和当前区间有交叉 或者start小于当前区间
48                    
49                     start=min(start,intervals[i].start);
50                     //后续就不用再找start了
51                     isfindStart=true;
52                    
53                     //如果start和end没有和其他Interval有交叉
54                     if(end<intervals[i].start) addCur=true;
55                     else addCur=false;
56                 }
57             }
58            
59            
60             if(!isfindEnd)
61             {
62                 //需要继续找end
63                 if(end>intervals[i].end)
64                 {
65                    
66                     if(start<=intervals[i].end) addCur=false;
67  
68                     if(i==n-1)
69                     {
70                         isfindEnd=true;
71                        
72                         if(addCur) result.push_back(intervals[i]);
73                         result.push_back(Interval(start,end));
74                        
75                         continue;
76                     }
77                 }
78                 else
79                 {
80                     //end小于当前区间，此时找到了end
81                     end=end>=intervals[i].start?intervals[i].end:end;
82  
83                     //找到了end
84                     isfindEnd=true;
85                     if(end>=intervals[i].start) addCur=false;
86                    
87                     result.push_back(Interval(start,end));
88                     if(addCur) result.push_back(intervals[i]);
89                    
90                     continue;
91                 }
92             }
93            
94             if(addCur) result.push_back(intervals[i]);
95         }
96         return result;
97     }
98 };