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)

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class Solution
{
public:
    bool search(int A[], int n, int target) 
    {
        int l = 0, r = n - 1;
        while(l <= r)
        {
           int m = (l + r) / 2;
            if(A[m] == target)
                return true;
            if(A[l] == A[m])
            {
                for(int i = l; i < m; ++ i)
                    if(A[i] == target)
                        return true;
                l = m + 1;
            }
            else if(A[l] < A[m])
            {
                if(A[l] <= target && target < A[m])
                    r = m - 1;
                else
                    l = m + 1;
            }
            else
            {
                if(A[m] < target && target <= A[r])
                    l = m + 1;
                else
                    r = m - 1;
            }
        }
        
        return false;
    }
};