33.Search in Rotated Sorted Array
===

###### tags: `Medium`,`Array`,`Binary Search`

[33. Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)

### 題目描述

There is an integer array `nums` sorted in ascending order (with **distinct** values).

Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` `(1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index `3` and become `[4,5,6,7,0,1,2]`.

Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of* `target` *if it is in* `nums`, *or* `-1` *if it is not in* `nums`.

You must write an algorithm with `O(log n)` runtime complexity.

### 範例

**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

**Example 3**

Input: nums = [1], target = 0
Output: -1

**Constraints**:

* 1 <= `nums.length` <= 5000
* -10^4^ <= `nums[i]` <= 10^4^
* All values of `nums` are **unique**.
* `nums` is an ascending array that is possibly rotated.
* -10^4^ <= `target` <= 10^4^

### 解答
#### Javascript

function search(nums, target) {
  let min = 0;
  let max = nums.length - 1;

  while (max >= min) {
    const mid = (max + min) >> 1;
    if (nums[mid] === target) return mid;

    if (nums[mid] >= nums[min]) {
      if (target >= nums[min] && target < nums[mid]) {
        max = mid - 1;
      } else {
        min = mid + 1;
      }
    } else {
      if (target <= nums[max] && target > nums[mid]) {
        min = mid + 1;
      } else {
        max = mid - 1;
      }
    }
  }

  return -1;
}

> [name=Marsgoat][time=Aug 8, 2023]

#### c#

public class Solution {
    public int Search(int[] nums, int target) {
        bool inLeft = target >= nums[0];
        int left = 0;
        int right = nums.Length - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;   
            if (nums[mid] == target) return mid;

            if (inLeft && nums[mid] < nums[left]) {
                right = mid - 1;
            } else if (!inLeft && nums[mid] > nums[right]) {
                left = mid + 1;
            } else {
                if (nums[mid] > target) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
        }
        
        return -1;
    }
}

> [name=Jim][time=Aug 8, 2023]
### Reference

[回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)