使用合并排序对(名称)进行排序



对重复的字符串进行排序时遇到问题,

这是我的代码..

成功地对第一个数组进行了排序,但在第二个数组(带有重复字符串)中,它似乎没有有序输出,你能帮我跟踪我的代码中的问题吗?

import java.util.*;
public class NewClass {
    public static void main(String[] args) {
        String[] ClassOne = { "Kring", "Panda", "Soliel", "Darryl", "Chan", "Matang", "Jollibee.", "Inasal" };
        String[] ClassTwo = { "Minnie", "Kitty", "Madonna", "Miley", "Zoom-zoom", "Cristine", "Bubbles", "Ara", "Rose", "Maria" };
        String[] names = new String[ClassOne.length + ClassTwo.length];
        mergeSort(ClassOne);
        mergeSort(ClassTwo);
        merge(names, ClassOne, ClassTwo);
        mergeSort(names);
        //Arrays.sort(names);
        for (String ClassThree : names) {
            System.out.println(ClassThree);
        }
    }
    public static void mergeSort(String[] names) {
        if (names.length > 2) {
            String[] left = new String[names.length / 2];
            String[] right = new String[names.length - names.length / 2];
            for (int i = 0; i < left.length; i++) {
                left[i] = names[i];
            }
            for (int i = 0; i < right.length; i++) {
                right[i] = names[i + names.length / 2];
            }
            mergeSort(left);
            mergeSort(right);
            merge(names, left, right);
        }
    }
    public static void merge(String[] names, String[] left, String[] right) {
        int a = 0;
        int b = 0;
        for (int i = 0; i < names.length; i++) {
            if (b >= right.length || (a < left.length && left[a].compareToIgnoreCase(right[b]) < 0)) {
                names[i] = left[a];
                a++;
            } else {
                names[i] = right[b];
                b++;
            }
        }
    }
}

下面是输出:

Ara
Chan
Cristine
Bubbles
Jollibee.
Inasal
Kring
Madonna
Matang
Miley
Minnie
Kitty
Panda
Rose
Maria
Soliel
Darryl
Zoom-zoom

更改

if (names.length > 2) {

if (names.length >= 2) {

输出

Ara
Bubbles
Chan
Cristine
Darryl
Inasal
Jollibee.
Kitty
Kring
Madonna
Maria
Matang
Miley
Minnie
Panda
Rose
Soliel
Zoom-zoom

只需更改此内容:

if (names.length > 2)

if (names.length > 1)

mergeSort递归运行,将数组分成两半,然后合并它们并返回到调用链上。当传递给mergeSort的数组长度为 <= 1 时,它认为数组已排序,这称为基本情况

为什么不将两个数组(String ClassOne 和 ClassTwo)连接成一个数组,然后在一个数组上调用 MergeSort?在我看来,您的过程使程序更加模棱两可。

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