【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 };