leetcode-905-easy

Sort Array By Parity

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]

思路一： 刚开始题目理解错了，错误理解为结果要有序，不过结果有序也能通过测试用例

    public int[] sortArrayByParity(int[] nums) {
        int[] clone = nums.clone();

        int begin = 0;
        int end = nums.length - 1;
        for (int i = 0; i < clone.length; i++) {
            if (clone[i] % 2 == 0) {
                nums[begin] = clone[i];
                begin++;
            }
        }

        for (int i = clone.length - 1; i >= 0; i--) {
            if (clone[i] % 2 != 0) {
                nums[end] = clone[i];
                end--;
            }
        }

        return nums;
    }