public class Solution {
public int[] SortArrayByParity(int[] nums) {
var countEven = 0;
for(int i = 0; i < nums.Length; i++){
if(nums[i] % 2 == 0){
(nums[i], nums[countEven]) = (nums[countEven], nums[i]);
countEven++;
}
}
return nums;
}
}
LINQ
public class Solution {
public int[] SortArrayByParity(int[] nums) {
var evenNumbers = nums.Where(n => n % 2 == 0);
var oddNumbers = nums.Where(n => n % 2 == 1);
return evenNumbers.Concat(oddNumbers).ToArray();
}
}