## Search in Rotated Sorted Array

### 描述

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.

### 分析

   7 │
6   │
─────┼───────────
│         5
│       4
│     3
│   2
│ 1

         7 │
6   │
5     │
4       │
3         │
───────────┼───────────
│   2
│ 1


### 代码

// Search in Rotated Sorted Array
// Time Complexity: O(log n)，Space Complexity: 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;
}
};