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📜  Java程序以波形形式对数组进行排序

📅  最后修改于: 2022-05-13 01:55:07.435000             🧑  作者: Mango

Java程序以波形形式对数组进行排序

给定一个未排序的整数数组,将数组排序为波状数组。如果 arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4] >= ... 则数组 'arr[0..n-1]' 以波形排序..

例子:

Input:  arr[] = {10, 5, 6, 3, 2, 20, 100, 80}
 Output: arr[] = {10, 5, 6, 2, 20, 3, 100, 80} OR
                 {20, 5, 10, 2, 80, 6, 100, 3} OR
                 any other array that is in wave form

 Input:  arr[] = {20, 10, 8, 6, 4, 2}
 Output: arr[] = {20, 8, 10, 4, 6, 2} OR
                 {10, 8, 20, 2, 6, 4} OR
                 any other array that is in wave form

 Input:  arr[] = {2, 4, 6, 8, 10, 20}
 Output: arr[] = {4, 2, 8, 6, 20, 10} OR
                 any other array that is in wave form

 Input:  arr[] = {3, 6, 5, 10, 7, 20}
 Output: arr[] = {6, 3, 10, 5, 20, 7} OR
                 any other array that is in wave form
 

一个简单的解决方案是使用排序。首先对输入数组进行排序,然后交换所有相邻元素。
例如,让输入数组为 {3, 6, 5, 10, 7, 20}。排序后,我们得到 {3, 5, 6, 7, 10, 20}。交换相邻元素后,我们得到 {5, 3, 7, 6, 20, 10}。

下面是这种简单方法的实现。

Java
// Java implementation of naive method for sorting
// an array in wave form.
import java.util.*;
  
class SortWave
{
    // A utility method to swap two numbers.
    void swap(int arr[], int a, int b)
    {
        int temp = arr[a];
        arr[a] = arr[b];
        arr[b] = temp;
    }
  
    // This function sorts arr[0..n-1] in wave form, i.e.,
    // arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4]..
    void sortInWave(int arr[], int n)
    {
        // Sort the input array
        Arrays.sort(arr);
  
        // Swap adjacent elements
        for (int i=0; i


Java
// A O(n) Java program to sort an input array in wave form
class SortWave
{
    // A utility method to swap two numbers.
    void swap(int arr[], int a, int b)
    {
        int temp = arr[a];
        arr[a] = arr[b];
        arr[b] = temp;
    }
  
    // This function sorts arr[0..n-1] in wave form, i.e.,
    // arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4]....
    void sortInWave(int arr[], int n)
    {
        // Traverse all even elements
        for (int i = 0; i < n; i+=2)
        {
            // If current even element is smaller
            // than previous
            if (i>0 && arr[i-1] > arr[i] )
                swap(arr, i-1, i);
  
            // If current even element is smaller
            // than next
            if (i


输出:

2 1 10 5 49 23 90

如果使用 O(nLogn) 排序算法(如合并排序、堆排序等),则上述解决方案的时间复杂度为 O(nLogn)。
这可以通过对给定数组进行一次遍历在 O(n) 时间内完成。这个想法是基于这样一个事实,即如果我们确保所有偶数定位(在索引 0、2、4 ……)的元素都大于它们相邻的奇数元素,我们就不需要担心奇数定位元素。以下是简单的步骤。
1)遍历输入数组的所有偶数定位元素,并执行以下操作。
....a) 如果当前元素小于前一个奇数元素,则交换前一个元素和当前元素。
....b)如果当前元素小于下一个奇数元素,则交换下一个和当前元素。

以下是上述简单算法的实现。

Java

// A O(n) Java program to sort an input array in wave form
class SortWave
{
    // A utility method to swap two numbers.
    void swap(int arr[], int a, int b)
    {
        int temp = arr[a];
        arr[a] = arr[b];
        arr[b] = temp;
    }
  
    // This function sorts arr[0..n-1] in wave form, i.e.,
    // arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4]....
    void sortInWave(int arr[], int n)
    {
        // Traverse all even elements
        for (int i = 0; i < n; i+=2)
        {
            // If current even element is smaller
            // than previous
            if (i>0 && arr[i-1] > arr[i] )
                swap(arr, i-1, i);
  
            // If current even element is smaller
            // than next
            if (i

输出:

90 10 49 1 5 2 23
 

请参阅完整的文章以波形排序数组以获取更多详细信息!