Question

这是我的要求：

使用2来创建（硬编码）50个州及其首府城市   维数组。

在对话框中：要求用户输入州或城市。

如果输入州，请找到其首府。如果进入城市，   找到它的国家。

如果没有，找到，发出错误消息。

这应该是循环，直到用户不想玩   了。

我真的不知道从哪里开始，到目前为止我所做的就是创建阵列，我真的不知道如何搜索阵列，并吐出相应的州/资本。 / p>

非常感谢任何帮助。

这是我到目前为止编写的代码。

import java.util.Scanner;

public class GuessStates {
    public static void main(String[] args){

          java.util.Scanner input = new java.util.Scanner(System.in);

     String[][] statesAndCapitols = {
        {"Alabama","Alaska","Arizona","Arkansas","California","Colorado","Connecticut","Delaware","Florida","Georgia","Hawaii","Idaho","Illinois","Indiana","Iowa","Kansas","Kentucky","Louisiana","Maine","Maryland","Massachusetts","Michigan","Minnesota","Mississippi","Missouri","Montana","Nebraska","Nevada","New Hampshire","New Jersey","New Mexico","New York","North Carolina","North Dakota","Ohio","Oklahoma","Oregon","Pennsylvania","Rhode Island","South Carolina","South Dakota","Tennessee","Texas","Utah","Vermont","Virginia","Washington","West Virginia","Wisconsin","Wyoming"},
        {"Montgomery","Juneau","Phoenix","Little Rock","Sacramento","Denver","Hartford","Dover","Tallahassee","Atlanta","Honolulu","Boise","Springfield","Indianapolis","Des Moines","Topeka","Frankfort","Baton Rouge","Augusta","Annapolis","Boston","Lansing","St. Paul","Jackson","Jefferson City","Helena","Lincoln","Carson City","Concord","Trenton","Santa Fe","Albany","Raleigh","Bismarck","Columbus","Oklahoma City","Salem","Harrisburg","Providence","Columbia","Pierre","Nashville","Austin","Salt Lake City","Montpelier","Richmond","Olympia","Charleston","Madison","Cheyenne"}};

            System.out.println("Please enter a State or a capitol city.");
                String userInput = input.nextLine();



        }
    }

再次感谢！

Answer 1

尝试使用for循环搜索数组。

使用for循环，它会跟踪并更新遍历数组的当前位置。

一旦找到正确的状态或资本（通过检查userInput.equalsIgnoreCase（statesAndCapitols [x] [y]），然后获取您所在的当前位置并检索所需的信息。

即

for(int x = 0; x < 2; ++x) //loop through states the first time, capitols the second
  for(int y = 0; y < 50; ++y) //always 50, unless new states get added (obviously not a problem in this example, but useful to think about in future problems - YOUR DATA WILL ALMOST ALWAYS CHANGE.
   if(userInput.equalsIgnoreCase(statesAndCapitols[x][y])
      System.out.println(statesAndCapitols[x == 1 ? 0 : 1][y]);

在数组中，我做了x == 1？ 0：1。这是一个三元运算符，它的含义是如果x等于1，则使用值0，否则使用值1.

这是解决这个问题的一种方法。

另一种方法是为城市和州创建自己的类/数据类型，这样您就不需要保持数组同步，这意味着您不需要更新2一个更改的项目（如添加另一个城市/州组合）。

希望这有点帮助！：）

Answer 2

String entered_state=input.nextLine();
for(int i=0;i<50;i++){
    if(statesAndCapitols[0][i].equals(entered_state)){
        String searched_city=statesAndCapitols[1][i];
        //print the city name
        break;
        }
}
if(i==50) 
//print error

从进入的城市搜索状态也是一样。

州和首都2d数组，java

2 个答案: