使用排序的的compareTo字符串数组的ArrayList（）方法 [英] Sorting an ArrayList of String arrays using compareTo() method

问题描述

我是新来的Java。

我想用排序字符串数组的ArrayList String.compareTo（）。

我编了code，使输出是：

  

因果关系，就是一个，关系

  
  

这是也，叫，因果关系

  
  

这是约一个，原因，以及它的，影响

现在我要排序的code（字典序），因此输出是：

  

因果关系，一个是，关系

  
  

同时，它也被称为，因果关系，是

  
  

据，一个约，影响和，原因是，其

不过，我产生了一些疯狂输出。

我的code如下。

任何帮助将是AP preciated。

我对这个可能很简单，几个小时的问题工作，我准备摧毁我的电脑。谢谢

 公共类Wk5Q5 {    无效过程1（）{        串S =因果关系是有关系的;
        字符串s2 =它也被称为因果关系;
        串S =这是关于一个原因及其影响;
        ArrayList的＆LT;的String []＆GT;名单=新的ArrayList＆LT;的String []＆GT;（）;
        的String [] = ARR1 s1.split（）;
        list.add（ARR1）;
        的String [] = ARR2 s2.split（）;
        list.add（ARR2）;
        的String [] = ARR3 s3.split（）;
        list.add（ARR3）;        / **
         * previously排序字符串数组，这样的数组列表
         *每个字是用逗号分隔
         * /
        的for（int i = 0; I＆LT;则为list.size（）;我++）{
            对于（INT J = 0; J＆LT; list.get（I）。长度; J ++）{
                串T = list.get（一）[J]。                如果（J＆0）{
                    T =，+ T;
                }
                System.out.print（T）;
                //System.out.println(list.get(i)[j]）;            }
            的System.out.println（）;
        }        / **
         *我试图在每个列表中的每个字符串排序
         * /
        对于（INT Z = 0; z，其中，则为list.size（）; Z ++）{
            的for（int i = 0; I＆LT; list.get（Z）。长度;我++）{
                字符串x = list.get（Z）[我]
                对于（INT J = I + 1; J＆LT; list.get（Z）。长度; J ++）{
                    字符串Y = list.get（Z）[J]。
                    如果（y.compareTo（X）小于0）{
                        字符串TEMP = list.get（Z）[我]
                        X = list.get（Z）[J]。
                        Y =温度;
                    }
                    System.out.print（x）的;
                }            }
        }
    }
 

解决方案

与实施 <问题EM>选择排序的算法是，你不要修改被排序的名单。当你换 X 和是，在列表中的相应位置的元素留在自己的老地方。

如果您停止使用 X 和是和 list.get代替其使用（Z）[我] 和 list.get（Z）[J] ，你的排序算法会产生不同的结果。更妙的是，如果作业允许您使用标准库，看看在<一个href=\"http://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#sort%28java.lang.Object%5B%5D%29\"相对=nofollow>内置Java中数组排序方式。

I am new to java.

I'm trying to sort an ArrayList of String arrays using String.compareTo().

I have compiled the code so that the output is:

Causality,is,a,relationship

It,is,also,called,causation

It,is,about,a,cause,and,its,affect

Now I want to sort that code (lexicographically) so the output is:

Causality,a,is,relationship

It,also,called,causation,is

It,a,about,affect,and,cause,is,its

However I am generating some crazy output.

My code is below.

Any help would be appreciated.

I have worked on this probably very simple problem for hours and I am ready to destroy my computer. Thanks

public class Wk5Q5 {

    void process1 () {

        String s1 = "Causality is a relationship";
        String s2 = "It is also called causation";
        String s3 = "It is about a cause and its affect";


        ArrayList<String[]> list = new ArrayList<String[]>();


        String[] arr1 = s1.split(" ");
        list.add(arr1);
        String[] arr2 = s2.split(" ");
        list.add(arr2);
        String[] arr3 = s3.split(" ");
        list.add(arr3);

        /**
         * previously sorted the arraylist of string arrays so that
         * each word is separated by commas
         */
        for(int i = 0; i < list.size(); i++){
            for (int j = 0; j < list.get(i).length; j++){
                String t = list.get(i)[j];

                if (j > 0){
                    t = ", " + t;   
                }
                System.out.print(t);
                //System.out.println(list.get(i)[j]);

            }
            System.out.println();
        }

        /**
         * my attempt at sorting each string in each list 
         */
        for(int z = 0; z < list.size(); z++){
            for(int i = 0; i < list.get(z).length; i++){
                String x = list.get(z)[i];
                for (int j = i+1; j < list.get(z).length; j++){
                    String y = list.get(z)[j];
                    if(y.compareTo(x) < 0) {
                        String temp = list.get(z)[i];
                        x = list.get(z)[j];
                        y = temp;
                    }
                    System.out.print(x);
                }

            }
        }
    }

解决方案

The problem with your implementation of the selection sort algorithm is that you do not modify the list being sorted. When you swap x and y, the elements in the corresponding positions of the list remain in their old places.

If you stop using x and y and replace their use with list.get(z)[i] and list.get(z)[j], your sorting algorithm would produce different results. Better yet, if the homework allows you to use standard library, take a look at a built-in way of sorting arrays in Java.

这篇关于使用排序的的compareTo字符串数组的ArrayList（）方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！