【LeetCode】81. Search in Rotated Sorted Array II

题目：

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.

 这题跟上一题一样，都是二分查找法，但是需要思考，对于时间复杂度，存在重复元素会有影响吗？可以考虑如[1,0,1,1,...,1]这样的数组，时间复杂度最坏会 变成O（N）。

class Solution 
{
public:
    int search(vector<int>& nums, int target)
    {
       int left = 0, right = nums.size()-1;
       while(left <= right)
       {
           int middle = left + ((right-left)>>1);
           if (nums[middle] == target) return true;
           if (nums[middle] < nums[right])
           {
               if (nums[middle] < target && target <= nums[right])left = middle + 1;
               else  right = middle - 1;
           }
           else if (nums[middle] > nums[right])
           {
               if (nums[left] <= target && target < nums[middle])right = middle - 1;
               else left = middle + 1;
           }
           else  right--;
       }
       return false;
    }
};