Force deduction solution summary 905- array sorted by parity

Directory links ：

Force buckle programming problem - The solution sums up _ Share + Record -CSDN Blog

GitHub Synchronous question brushing items ：

https://github.com/September26/java-algorithms

Original link ： Power button

describe ：

Give you an array of integers nums, take nums Move all even elements in the array to the front of the array , Followed by all odd elements .

Returns... That meets this condition Any array As the answer .

Example 1：

Input ：nums = [3,1,2,4]
Output ：[2,4,3,1]
explain ：[4,2,3,1]、[2,4,1,3] and [4,2,1,3] It will also be seen as the right answer .
Example 2：

Input ：nums = [0]
Output ：[0]
 

Tips ：

1 <= nums.length <= 5000
0 <= nums[i] <= 5000

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/sort-array-by-parity
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Their thinking ：

*  Their thinking ：
*  Traverse twice , Find the even number for the first time , Find the odd number for the second time 

Code ：

public class Solution905 {

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


}