LeetCode in Kotlin
905. Sort Array By Parity
Easy
Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.
Return any array that satisfies this condition.
Example 1:
Input: nums = [3,1,2,4]
Output: [2,4,3,1]
Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
Example 2:
Input: nums = [0]
Output: [0]
Constraints:
1 <= nums.length <= 50000 <= nums[i] <= 5000
Solution
class Solution {
fun sortArrayByParity(nums: IntArray): IntArray {
var temp: Int
var i = 0
for (k in nums.indices) {
if (nums[k] % 2 == 0) {
temp = nums[k]
nums[k] = nums[i]
nums[i] = temp
i++
}
}
return nums
}
}