81. Search in Rotated Sorted Array II

该题与Search in Rotated Sorted Array 同理，由于存在重复元素，当mid处与边缘right位置处的值相同时，若边缘位置不为target，则边缘right左移一位：

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int left=0,right=nums.size()-1;
        int mid=(right+left)/2;
        while(left<=right)
        {
            if(nums[mid]==target/*||nums[left]==target||nums[right]==target*/)
                return 1;
            else
            {
                if(nums[mid]<nums[right]) //right is ordered
                {
                    if(target>nums[mid]&&target<=nums[right])
                    {
                        left=mid+1;
                    }
                    else
                    {
                        right=mid-1;
                    }
                    
                }
                else //left is ordered
                {
                    if(nums[left]<nums[mid])
                    {                      
                        if(target>=nums[left]&&target<nums[mid])
                            right=mid-1;
                        else
                            left=mid+1;
                    }
                    else
                    {
                      if(nums[right]==target)
                          return 1;
                        else
                            right--;
                    }
                }
                
            }
            
            mid =(left+right)/2;
            
        }
        
        return 0;
    }
};