leetcode 905. Sort Array By Parity（按奇偶排序数组）

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.

按奇偶顺序排序，把数组的偶数放在前面，奇数放在后面。

思路：
遍历两次数组，第一次把偶数填进前半段，第二次把奇数填到后半段。

public int[] sortArrayByParity(int[] nums) {
    int n = nums.length;
    int[] res = new int[n];
    int i = 0;
    
    for(int num : nums) {
        if(num % 2 == 0) {
            res[i] = num;
            i ++;
        }
    }
    
    for(int num : nums) {
        if(num % 2 != 0) {
            res[i] = num;
            i ++;
        }
    }
    return res;
}