LeetCode in Kotlin
912. Sort an Array
Medium
Given an array of integers nums, sort the array in ascending order and return it.
You must solve the problem without using any built-in functions in O(nlog(n)) time complexity and with the smallest space complexity possible.
Example 1:
Input: nums = [5,2,3,1]
Output: [1,2,3,5]
Explanation: After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
Example 2:
Input: nums = [5,1,1,2,0,0]
Output: [0,0,1,1,2,5]
Explanation: Note that the values of nums are not necessairly unique.
Constraints:
1 <= nums.length <= 5 * 104-5 * 104 <= nums[i] <= 5 * 104
Solution
class Solution {
fun sortArray(nums: IntArray): IntArray {
return mergeSort(nums, 0, nums.size - 1)
}
private fun mergeSort(arr: IntArray, lo: Int, hi: Int): IntArray {
if (lo == hi) {
val sortedArr = IntArray(1)
sortedArr[0] = arr[lo]
return sortedArr
}
val mid = (lo + hi) / 2
val leftArray = mergeSort(arr, lo, mid)
val rightArray = mergeSort(arr, mid + 1, hi)
return mergeSortedArray(leftArray, rightArray)
}
private fun mergeSortedArray(a: IntArray, b: IntArray): IntArray {
val ans = IntArray(a.size + b.size)
var i = 0
var j = 0
var k = 0
while (i < a.size && j < b.size) {
if (a[i] < b[j]) {
ans[k++] = a[i++]
} else {
ans[k++] = b[j++]
}
}
while (i < a.size) {
ans[k++] = a[i++]
}
while (j < b.size) {
ans[k++] = b[j++]
}
return ans
}
}