数组（3）：Search in Rotated Sorted Array

describe:

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.

Analysis:
binary search method, the difficulty lies in the determination of the left and right boundaries

code

// LeetCode, Search in Rotated Sorted Array
// 时间复杂度 O(log n)，空间复杂度 O(1)
class Solution {
public:
int search(const vector<int>& nums, int target) {
int first = 0, last = nums.size();
while (first != last) {
const int mid = first + (last - first) / 2;
if (nums[mid] == target)
return mid;
if (nums[first] <= nums[mid]) 
{
  if (nums[first] <= target && target < nums[mid])
      last = mid;
  else
      first = mid + 1;
} 
else 
{
  if (nums[mid] < target && target <= nums[last-1])
      first = mid + 1;
  else
      last = mid;
}
}
return -1;
}
};