Bayi. Search in rotated Sorted Array II

Follow to "Search in rotated Sorted Array":
What if duplicates is allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given the target was in the array.

This topic is in https://leetcode.com/problems/search-in-rotated-sorted-array/
On the basis of the addition of the Allow to have the same element for this condition.
Then there will be a special case, such as 1,1,1,1,1,3,3 after rotation for 1,3,3,1,1,1,1.
When the middle element equals the first element, put l++, because target! =nums[mid].
and Nums[mid]=nums[l], can be launched target! =NUMS[L].

classSolution { Public:BOOLSearch vector<int>& Nums,intTarget) {intL=0, R=nums.size ()-1; while(L&LT;=R) {intMid= (L+R)/2;if(Nums[mid]==target)return true;Else if(Nums[mid]>nums[l]) {//increments the sequence on the left                if(Target>=nums[l]&&target<nums[mid]) r=mid-1;ElseL=mid+1; }Else if(Nums[mid]<nums[l]) {if(Target>nums[mid]&&target<=nums[r]) l=mid+1;Elser=mid-1; }Else if(Nums[mid]==nums[l])            {l++; }        }return false; }};

and Search in rotated Sorted Array
Suppose a sorted array is rotated on some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You is given a target value to search. If found in the array is return its index, otherwise return-1.
This topic only need to change the parameters returned to AC.

classSolution { Public:intSearch vector<int>& Nums,intTarget) {intL=0, R=nums.size ()-1; while(L&LT;=R) {intMid= (L+R)/2;if(Nums[mid]==target)returnMidElse if(Nums[mid]>nums[l]) {if(Target>=nums[l]&&target<nums[mid]) r=mid-1;ElseL=mid+1; }Else if(Nums[mid]<nums[l]) {if(Target>nums[mid]&&target<=nums[r]) l=mid+1;Elser=mid-1; }Else if(Nums[mid]==nums[l])            {l++; }        }return-1; }};

Bayi. Search in rotated Sorted Array II