题目链接：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.

这道题的要求是在Search in Rotated Sorted Array的基础上允许数组中出现重复元素。

这道题是Search in Rotated Sorted Array的扩展，允许数组中出现重复元素。

思路类似，只不过当A[l] == A[m]的时候，无法判断左侧是否旋转，因此需要遍历l到m之间的元素进行查找，这样，最差的时间复杂度为O(n)。

时间复杂度：O(logn)

空间复杂度：O(1)

 1 class Solution 2 { 3 public: 4     bool search(int A[], int n, int target)  5     { 6         int l = 0, r = n - 1; 7         while(l <= r) 8         { 9            int m = (l + r) / 2;10             if(A[m] == target)11                 return true;12             if(A[l] == A[m])13             {14                 for(int i = l; i < m; ++ i)15                     if(A[i] == target)16                         return true;17                 l = m + 1;18             }19             else if(A[l] < A[m])20             {21                 if(A[l] <= target && target < A[m])22                     r = m - 1;23                 else24                     l = m + 1;25             }26             else27             {28                 if(A[m] < target && target <= A[r])29                     l = m + 1;30                 else31                     r = m - 1;32             }33         }34         35         return false;36     }37 };