題目描述：
Suppose an array sorted in ascending order 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.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
程序：
class Solution {
public:
int search(vector<int>& nums, int target) {
if(nums.size() == 0) return -1;
if(nums.size() == 1 && nums[0] == target) return 0;
int i = 0, j = nums.size()-1;
while(i < j){
if(nums[i] < target) i++;
if(nums[j] > target) j--;
if(nums[i] == target) return i;
if(nums[j] == target) return j;
if(nums[i] > target && nums[j] < target) return -1;
}
return -1;
}
};