LeetCode Online Judge 题目C# 练习 - 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.

 1         public static bool SearchinRotatedSortedAarrayII(int[] A, int target)

 2         {

 3             return BinarySearchRotatedSortedArrayII(A, 0, A.Length - 1, target);

 4         }

 5 

 6         public static bool BinarySearchRotatedSortedArrayII(int[] A, int l, int r, int target)

 7         {

 8             if (l > r)

 9                 return false;

10 

11             int m = (l + r) / 2;

12 

13             if (target == A[m])

14                 return true;

15 

16             // if duplicate is allowed, we have to search both upper and lower halves

17             // corner case : 8,9,5,8,8,8,8,8 and 8,8,8,8,9,5,8

18             if (A[m] == A[l] && A[m] == A[r])

19             {

20                 return BinarySearchRotatedSortedArrayII(A, l, m - 1, target) ||

21                         BinarySearchRotatedSortedArrayII(A, m + 1, r, target);

22             }

23 

24             if (A[m] >= A[l])

25             {

26                 if (target >= A[l] && target < A[m])

27                     r = m - 1;

28                 else

29                     l = m + 1;

30 

31                 return BinarySearchRotatedSortedArrayII(A, l, r, target);

32             }

33             else

34             {

35                 if (target > A[m] && target <= A[r])

36                     l = m + 1;

37                 else

38                     r = m - 1;

39 

40                 return BinarySearchRotatedSortedArrayII(A, l, r, target);

41             }

42 

43         }

代码分析：

　　如注释里说的，如果允许重复，那就要前半部后半部都要找。 我承认，我掉入陷阱里了！！！ 如果数组全是 0 ， 要你找1. 那么时间复杂度是多少？O(n)吧。既然是O(n)，干嘛不就loop一遍呢？

1         public static bool SearchinRotatedSortedAarrayIIOpt(int[] A, int target)

2         {

3             for (int i = 0; i < A.Length; i++)

4             {

5                 if (A[i] == target)

6                     return true;

7             }

8             return false;

9         }