##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?
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
Write a function to determine if a given target is in the array.
The array may contain duplicates.
##思路
这道题目，如果仅仅是实现，是很简单的，最坏情况O(n)，我们知道在有序数组的情况下，二分搜索的时间复杂度是O(logn)，那么在存在反转的情况下，能否使用二分搜索呢？答案是肯定的，唯一的区别就是三个索引值确定需要修改：left,right,mid
##代码
class Solution {
public:
bool search(vector<int>& nums, int target) {
if(nums.empty())
return false;
int l=0,r=nums.size()-1;
int mid = 0;
while(l<r)
{
mid = (l+r)/2;
if(nums[mid]==target)//如果相等
return true;
if(nums[mid]>nums[r])//如果大于
{
if(nums[mid]>target&&nums[l]<=target)
r=mid-1;
else
l=mid+1;
}
else if(nums[mid]<nums[r])//如果小于
{
if(nums[mid]<target&&nums[r]>=target)
l=mid+1;
else
r=mid-1;
}
else //如果等于
r--;
}
return nums[l]==target?true:false;
}
};