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.
在循环错位的有序数组中进行搜索，与33题类似
https://cloud.tencent.com/developer/article/1019304
不同的是元素可能重复，
当遇到等于的时候，只能一个个地缩小区间，而不能以log复杂度减小，所以在最坏情况下，效率会退化为O(n).
代码语言：javascript复制class Solution {
public:
bool search(vector<int>& nums, int target) {
int l=0,r=nums.size()-1;
while(l<=r)
{
int mid=(l r)>>1;
if(nums[mid]==target) return true;
if(nums[mid]<nums[l])
{
if(target<nums[l] && target>nums[mid])
l=mid 1;
else
r=mid-1;
}
else if(nums[mid]>nums[l])
{
if(target>=nums[l] && target<nums[mid])
r=mid-1;
else
l=mid 1;
}
else
l ; //mid和l值相等，无法获知mid的位置
}
return false;
}
};
