标签：abs   抛出异常   str   输入   一个   奇数   form   OLE   format   

前提条件

1. 时间复杂度：O(n)
2. 空间复杂度：O(1)

思路

1. 若数组长度为偶数，那么其中奇数元素和偶数元素的个数相等；若数组长度为奇数，那么其中奇数元素和偶数元素的个数相差 1；我们先校验奇数元素个数和偶数元素个数是否合法，若不合法，则直接抛出异常；
2. 根据数组中奇数元素的个数和偶数元素的个数的大小来判断结果数组中第一个元素应该是奇数还是偶数；
3. 感觉代码写的很丑，暂时想不到更加优雅的写法了。。。

import java.util.Arrays;

class Solution {

    public static void main(String[] args) {
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{2, 2, 3})));
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{1, 2, 3, 4})));
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{1, 2, 3, 4, 4})));
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{1, 2, 3, 4, 5})));
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{1, 2, 3, 4, 5, 6})));
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{1, 2, 3, 4, 5, 6, 7})));
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{1, 2, 3, 4, 5, 6, 7, 8})));
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{1, 2, 3, 4, 5, 6, 7, 8, 8})));
        System.out.println(Arrays.toString(oddEvenHandle(new int[]{1, 2, 3, 4, 5, 6, 7, 8, 8, 8})));
    }

    private static int[] oddEvenHandle(int[] array) {
        if (array == null || array.length == 0) {
            return array;
        }

        int oddCount = 0, evenCount = 0;
        for (int ele : array) {
            if (ele % 2 == 0) {
                evenCount++;
            } else {
                oddCount++;
            }
        }

        if (Math.abs(oddCount - evenCount) > 1) {
            throw new IllegalArgumentException(String.format("oddCount = %d, evenCount = %d", oddCount, evenCount));
        }

        boolean oddHead = false;
        if (oddCount >= evenCount) {
            oddHead = true;
        }

        int oddPoint = 0, evenPoint = 0;
        for (int i = 0; i 

输入一个整数数组，输出奇偶数相间排列的数组

标签：abs   抛出异常   str   输入   一个   奇数   form   OLE   format   

原文地址：https://www.cnblogs.com/optor/p/13185029.html