接受 5 个名字并打印最长的名字？

Question

我想要 5 个名字中最长的名字。我想我应该使用compareTo() 方法还是length()？

输出必须是这样的：

enter 5 names :
Joey
Mark
Catherine
Zachery
Foster

Longest name is Catherine.

我应该使用什么方法以及如何使用？到目前为止，这是我的代码：

Scanner x = new Scanner(System.in);
String name = ""
System.out.print("Enter 5 names");
name = x.nextLine();
name2 = x.nextLine();
name3 = x.nextLine();
name4 = x.nextLine();
name5 = x.nextLine();

if(name.compareTo(name2)>0)  //is this method right?

Answer 1

.compareTo 告诉您哪个字符串按字典顺序排在第一位（0 如果 s1>s2）

String s1 = "abc";
String s2 = "def";
s1.compareTo(s2) < 0;

.length() 返回字符串的长度

s1.length()==3;

在您的情况下，您需要根据长度进行比较，因此您需要后者。如果只有 5 个名字，你可以取第一个并假设它是最长的，然后通过保存“迄今为止最长的”并在它们出现时进行比较来逐一阅读其他的。毕竟，你只关心最长的。

如果您希望它们按长度排序，同时仍保留它们全部，则需要将它们存储在某种集合（列表、数组）中，然后根据长度对其进行排序。

问题很简单，我就不直接提供代码了，自己去摸一摸，可以的：）

Answer 2

import java.util.Scanner;

public class LenghtyName {

    public static void main(String[] args) {

        Scanner x = new Scanner(System.in);
        /*
         Instead of declaring 5 different String variable
         just use String array with size = 5
         */
        String[] names = new String[5];

        System.out.print("Enter 5 names :");

        names[0] = x.nextLine();
        names[1] = x.nextLine();
        names[2] = x.nextLine();
        names[4] = x.nextLine();
        names[5] = x.nextLine();

        //Assume lenthyName as empty String
        String lengthyName = "";

        /*
         Iterate over String array using for-each loop
         */
        for (String name : names) {
            /*
             -Check null to avoid NullPointerException
             -Trim the left and right blank space in name by #trim()
             -Compare current name length with lengthyName if greater 
             replace the lengthyName by current name.
             */
            if (name != null && name.trim().length() > lengthyName.length()) {
                lengthyName = name;
            }
        }

        /*
         Print length name
         */
        System.out.println("Longest name is " + lengthyName);
    }
}

Answer 3

这个怎么样？

Scanner in = new Scanner(System.in);
// no need to have 5 all over the code.
// Define it once to avoid "magic nubmers"
int namesCount = 5;
// fun fact: shortest name is always "". We don't have to use null
String longestName = "";
System.out.print("Enter " + nameCount + " names:");
for (int i=0; i< nameCount; i++){
    // store new name from user
    String candidate = in.readLine();
    // is this one longer than the current longest?
    if (longestName.length() < candidate.length()){
        // found a longer name
        longestName = candidate;
    }
}
System.out.println("Longest name is " + longestName); 

这放弃了存储名称，因为您似乎只使用最长的名称。它还概括了要迭代的名称数量，最重要的是变量名称是有意义的名称。

Answer 4

这是一个可行的解决方案：

public class LongestWord {

    public static String getLongestString(String[] array) {
          int maxLength = 0;
          String longestString = null;
          for (String s : array) {
              if (s.length() > maxLength) {
                  maxLength = s.length();
                  longestString = s;
              }
          } 
          return longestString;
      }

      public static void main(String[] args) {
          String[] toppings = {"Cheeddddddddddse", "Pepperoni", "Black Olivesddd"};
          String longestString = getLongestString(toppings);
          System.out.format("longest string: '%s'\n", longestString);
      }

}

Answer 5

一个简单的方法是使用for 循环来读取 5 个名称并找到最大名称的长度。使用for 循环可避免创建 5 个威慑字符串变量。

如果您以后想使用这些名称，可以使用 String 数组。

public static void main(String[] args) throws IOException {
    Scanner x = new Scanner(System.in);
    String name = "";
    String maxName = "";
    int maxLength = 0;

    System.out.println("enter 5 name :");

    for (int i = 0; i < 5; i++) { // read 5 names
        name = x.nextLine();
        if (maxLength < name.length()) { // check longest name
            maxLength = name.length();
            maxName = name;              // store in temp variable to show
        }
    }
    System.out.println("Longest name is " + maxName); // print largest name
}

输出：

输入5个名字：
拉吉
西塔
歌
慕克吉
丽塔
最长的名字是慕克吉

Answer 6

这是一个适用于任意数量条目的解决方案：

Scanner x = new Scanner(System.in);
String name = "", temp="";

while (x.hasNextLine()){
    temp = x.nextLine();
    if (temp.length() > name.length()) name = temp;
}

System.out.println("Longest is " + name);

您需要Ctrl + Z 在 Windows 上结束 inputStream

Answer 7

这是我用来比较 3 个输入字符串的长度的代码：

public static void main(String[] args) {

    String string1 , string2 , string3;

    Scanner input = new Scanner(System.in);
    System.out.println("Enter three string names to compare");
    string1 = input.nextLine();
    string2 = input.nextLine();
    string3 = input.nextLine();

    if (string1.length()>string2.length()) {
        if (string1.length() > string3.length())
            System.out.println("String 1 has the longest length , length = "+string1.length());
    }

    if (string2.length()>string1.length()){
        if(string2.length()>string3.length())
            System.out.println("String 2 has the longest length , length = "+string2.length());
    }

    if (string3.length()>string1.length()) {
        if (string3.length() > string2.length())
            System.out.println("String 3 has the longest length , length = " + string3.length());
    }

}

Answer 8

//SIMPLE JAVA SOLUTION
class GFG 
{
    String longest(String names[], int n)
 {
        String res="";
        for(int i=0;i<n;i++)
        {
            if(res.length()<names[i].length())
            {
                res=names[i];
            }
        }
        return res;
        
    }
}