Search in Rotated Sorted Array

## Description

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
``````

## Solution

``````// 递归写法
func search(nums []int, target int) int {
if len(nums) == 0 {
return -1
}

mid := len(nums) / 2

if nums[mid] == target {
return mid
}

search_right := func() int {
r := search(nums[mid:], target)
if r == -1 {
return -1
}
return mid + r
}

search_left := func() int {
return search(nums[:mid], target)
}

// 判断左边是否有序
if nums[0] < nums[mid] {
// 左边有序时判断是否在左边范围
if target >= nums[0] && target <= nums[mid] {
return search_left()
}

return search_right()
} else {
// 右边有序时判断是否在右边范围
if target >= nums[mid] && target <= nums[len(nums) - 1] {
return search_right()
}

return search_left()
}
}``````