LeetCode in Kotlin
33. Search in Rotated Sorted Array
Medium
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.
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 <= 5000-104 <= nums[i] <= 104- All values of
numsare unique. numsis an ascending array that is possibly rotated.-104 <= target <= 104
Solution
class Solution {
fun search(nums: IntArray, target: Int): Int {
var mid: Int
var lo = 0
var hi = nums.size - 1
while (lo <= hi) {
mid = (hi - lo shr 1) + lo
if (target == nums[mid]) {
return mid
}
// if this is true, then the possible rotation can only be in the second half
if (nums[lo] <= nums[mid]) {
// the target is in the first half only if it's
if (nums[lo] <= target && target <= nums[mid]) {
// included
hi = mid - 1
} else {
// between nums[lo] and nums[mid]
lo = mid + 1
}
// otherwise, the possible rotation can only be in the first half
} else if (nums[mid] <= target && target <= nums[hi]) {
// the target is in the second half only if it's included
lo = mid + 1
} else {
// between nums[hi] and nums[mid]
hi = mid - 1
}
}
return -1
}
}