java - 在java中查找带领带的最长名字

Question

public void longestName(Scanner console, int num) {
    String longest = "";
    boolean tie = false;

    for(int i = 1; i <= num; i++) {
        System.out.print("name #" + i + "? " );
        String name = console.next();

        if(name.length() == longest.length()) {
            tie = true;
        } else if(name.length() > longest.length()) {
            tie = false;
            longest = name;
        }
    }

    String capitalized = longest.substring(0, 1).toUpperCase() +
                         longest.substring(1).toLowerCase();
    System.out.println(capitalized + "'s name is longest");

    if(tie)
        System.out.println("(There was a tie!)");
}}

if 循环始终返回 tie 为 true。如何仅检查最长姓名的领带？

Answer 1

仅当存在多个具有相同字符且最长的名称时，if 才会返回 tie 为 true。或者，领带是假的。
将代码更改为可运行的类，测试如下:

import java.util.Scanner;

public class Test {
    public static void main(String args[]){
        longestName(new Scanner(System.in), 3);
    }

    public static void longestName(Scanner console, int num) {

        String longest = "";

        boolean tie = false;

        for (int i = 1; i <= num; i++) {

            System.out.print("name #" + i + "? ");

            String name = console.next();

            if (name.length() == longest.length()) {

                tie = true;

            } else if (name.length() > longest.length()) {

                tie = false;

                longest = name;

            }

        }

        String capitalized = longest.substring(0, 1).toUpperCase() +

                longest.substring(1).toLowerCase();

        System.out.println(capitalized + "'s name is longest");

        if (tie)

            System.out.println("(There was a tie!)");
    }
}

输入为 jordan、tom 和 bob，输出如下:

name #1? jordan
name #2? tom
name #3? bob
Jordan's name is longest

另一个测试是:

name #1? mary
name #2? andy
name #3? tom
Mary's name is longest
(There was a tie!)