【leetcode】Search in Rotated Sorted Array II

Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

 

与 I类似，只是在二分搜索前，只是在判断左右两边是否有序时更加麻烦一些。

此时复杂度是O(n)了，和采用线性搜索的方法基本一致

 

 1 class Solution {

 2 public:

 3     bool search(int A[], int n, int target) {

 4        

 5        

 6         int left=0;

 7         int right=n-1;

 8         int mid;

 9         while(left<=right)

10         {

11            

12             mid=(left+right)/2;

13             if(A[mid]==target)

14             {

15                 return true;

16             }

17            

18  

19             //增加了这些判断语句

20             bool isLeft=true;

21            

22             for(int i=left;i<=mid;i++)

23             {

24                 if(A[i]>A[mid])

25                 {

26                     isLeft=false;

27                     break;

28                 }

29             }

30            

31             if(isLeft)

32             {

33                 if(A[left]<=target&&target<A[mid])

34                 {

35                     right=mid-1;

36                 }

37                 else

38                 {

39                     left=mid+1;

40                 }

41             }

42             else

43             {

44                 if(A[mid]<target&&target<=A[right])

45                 {

46                     left=mid+1;

47                 }

48                 else

49                 {

50                     right=mid-1;

51                 }

52             }

53  

54         }

55         return false;

56        

57     }

58 };