JZ-013-Adjust the array order so that odd numbers are in front of even numbers

Adjust the order of the array so that odd numbers come before even numbers

Title description

Enter an integer array and implement a function to adjust the order of the numbers in the array so that all odd numbers are in the first half of the array, and all even numbers are in the second half of the array, and the relative between odd and odd, even and even is guaranteed The location remains unchanged.

title link : adjust the array order so that the odd number is before the even number

Code

/**
 * 标题：调整数组顺序使奇数位于偶数前面
 * 题目描述
 * 输入一个整数数组，实现一个函数来调整该数组中数字的顺序，使得所有的奇数位于数组的前半部分，所有的偶数位于数组的后半部分，
 * 并保证奇数和奇数，偶数和偶数之间的相对位置不变。
 * 题目链接：
 * https://www.nowcoder.com/practice/beb5aa231adc45b2a5dcc5b62c93f593?tpId=13&&tqId=11166&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
 */
public class Jz13 {

    /**
     * 方法一：创建一个数组
     *
     * @param array
     */
    public void reOrderArray(int[] array) {
        // 奇数个数
        int oddCnt = 0;
        for (int x : array) {
            if (!isEven(x)) {
                oddCnt++;
            }
        }
        int[] copy = array.clone();
        int i = 0, j = oddCnt;
        for (int num : copy) {
            if (num % 2 == 1) {
                array[i++] = num;
            } else {
                array[j++] = num;
            }
        }
    }

    private boolean isEven(int x) {
        return x % 2 == 0;
    }

    /**
     * 方法二：使用冒泡思想，每次都把当前偶数上浮到当前最右边。时间复杂度 O(N2)，空间复杂度 O(1)，时间换空间。
     *
     * @param array
     */
    public void reOrderArray2(int[] array) {
        int N = array.length;
        for (int i = N - 1; i > 0; i--) {
            for (int j = 0; j < i; j++) {
                if (isEven(array[j]) && !isEven(array[j + 1])) {
                    swap(array, j, j + 1);
                }
            }
        }
    }

    private void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}

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