Search in Rotated Sorted Array II

描述
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.
中文：
如果允许重复元素，怎么办

分析：
重复元素的话，主要是
当nums[first]==nums[mid]时，无法判断元素的序列是否递增，也无法找到转折点如 131111111怎么找？
所以要单独分析，那就只能将first向前推进。

class Solution { public:        bool search(vector<int>& nums, int target) {        int first=0;        int last=nums.size()-1;        int mid;       while(first<=last)       {          mid=(first+last)/2;          if(nums[mid]==target)          return true;          if(nums[first]<nums[mid])          {           if(target>=nums[first]&&target<nums[mid])           last=mid-1;           else           first=mid+1;         }      else if(nums[first]>nums[mid])       {         if(target>nums[mid]&&target<=nums[last])         first=mid+1;         else         last=mid-1;     }     else      first++;    }    return false;}}; 

Runtime: 8 ms