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.

Solution:

唯一需要判断的地方是如果刚好在duplicate的地方分开了怎么办，这样的时候 mid没有意义了。

 1 public class Solution {

 2     public boolean search(int[] A, int target) {

 3         // IMPORTANT: Please reset any member data you declared, as

 4         // the same Solution instance will be reused for each test case.

 5         int end = A.length - 1;

 6         int start = 0;

 7         if (A[start] == A[end]){

 8             if (A[start]==target) return true;

 9             int i = A[start];

10             while (start <= end && A[start] == i) start ++;

11             while (start <= end && A[end] == i) end --;

12         }

13         while(start <= end){

14             int mid = (end + start) / 2;

15             if(target == A[mid]) return true;

16             if(A[start] <= A[mid]){//left half is sorted

17                 if(A[start] <= target &&  target < A[mid]) end = mid - 1;

18                 else start = mid + 1;

19             }

20             else{

21                 if(A[mid] < target && target <= A[end]) start = mid + 1;

22                 else end = mid - 1;

23             }

24         }

25         return false;

26     }

27 }