0

我试图让这个程序的 groupPairs 函数在初始字符串数组 [One,Two,Three,Four,Five,Six] 中获取六个字符串,并创建一个大小为原来的一半 (3) 的新字符串数组六个字符串按对 [OneTwo,ThreeFour,FiveSix] 顺序分组,然后将生成的新 String[] 返回给 main 方法。

import java.util.*;

public class Application
{
    static String[] groupPairs(String[] array)
    {
        String[] newArray = new String[(array.length)/2];
        int count=0;
        for(String string:newArray)
        {
            newArray[count]=array[count].append(array[count+1]);
            count=count+2;
        }
        return newArray;
    }
    public static void main(String args[]) //main method, don't worry about this
    {
        String[] list = new String[5];
        list[0]="One";
        list[1]="Two";
        list[2]="Three";
        list[3]="Four";
        list[4]="Five";
        list[5]="Six";
        String[] list2 = groupPairs(list);
    }
}

尝试编译程序时,出现此错误:

Application.java:11: cannot find symbol
symbol  : method append(java.lang.String)
location: class java.lang.String
            newArray[count]=array[count].append(array[count+1]);
                                        ^

任何关于如何修复这条线的想法,以便我的新数组将连接原始 String[] 值的对,将不胜感激!

4

1 回答 1

0

您不能对数组执行追加操作。试试下面的。

String[] list = new String[6];
list[0] = "One";
list[1] = "Two";
list[2] = "Three";
list[3] = "Four";
list[4] = "Five";
list[5] = "Six";
String[] list2 = new String[list.length / 2];
for (int i = 0, j = 0; i < list.length; i++, j++)
{
list2[j] = list[i] + list[++i];
}

于 2012-12-11T22:06:49.800 回答