Suppose a sorted array is rotated at 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 are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.

旋轉數(shù)組是分為兩個有序數(shù)組，因此可以使用二分查找。
每次判斷一段數(shù)組是否有序，再根據(jù)有序部分首尾兩個數(shù)字和target的大小關系來判斷target存在于哪一部分。有序則調用正常二分，找不到則對剩下一半數(shù)組調用此函數(shù)。

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        int res = -1, left = 0, right = n - 1;
        
        while(left <= right)
        {
            int mid = (left + right) >> 1;
            if(target == nums[mid])
            {
                    res = mid;
                    break;
            }
            else if(nums[mid] < nums[right]) //后半部分有序
            {
                if(target > nums[mid] && target <= nums[right])
                    left = mid + 1;
                else
                    right = mid - 1;
            }
            else //前半部分有序
            {
                if(target >= nums[left] && target < nums[mid])
                    right = mid - 1;
                else
                    left = mid + 1;
            }
        }    
        return res;
    }
};