(LeetCode) 33. Search in Rotated Sorted Array

33. Search in Rotated Sorted Array

• Explore : Interview > Top Interveiw Questions > Medium Collection
• 분류 : Sorting and Searching
• 난이도 : Medium

Problem

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

Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= 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 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.

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 guaranteed to be rotated at some pivot.
• -10^4 <= target <= 10^4

Solution

Exapnder

class Solution {
    fun search(nums: IntArray, target: Int): Int {
        if (nums.contains(target)) {
            return nums.indexOf(target)
        }
        return -1
    }
}

Point of Thinking

• nums에 target이 있으면 index를 반환한다.
• target이 없으면 -1을 반환하면 Accepted.