33. Search in Rotated Sorted Array

二分法：每次判断mid所在的数段是否有序

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