Question

我的代码检查一定数量的用户输入的字符串是否有任何重复的字符。例如，如果我输入字符串＆＃34; google＆＃34; ＆＃34;纸＆＃34;和＆＃34; water＆＃34;，代码返回＆＃34; paper＆＃34;和＆＃34; water＆＃34 ;;因为＆＃34; google＆＃34;有两个Os。

我将代码部分放下，但是在打印时，在输出的最后一个字符串之后会出现一个空格，我无法弄清楚如何摆脱它。

import java.util.Scanner;
import java.util.*;

class words{

    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter the number or words: ");
        String[] words = new String[sc.nextInt()];
        System.out.print("Enter the strings: ");
        boolean truth = false;

        for (int i = 0; i < words.length; i++) {
            words[i] = sc.next();
        }
        for(int i=0;i<words.length;i++){
            int j;
            for(j=1;j<words[i].length();j++) {
                if(words[i].charAt(j) == words[i].charAt(j-1)){
                    break;
                }
            }
            if(j==words[i].length()){
                truth = true;
                System.out.print(words[i]+" ");
            }
        }
        if(!truth){
            System.out.println("NONE");
        }
    }
}

Answer 1

功能使逻辑可读

移动逻辑以检查重复字符到函数中;我会利用NVIDIA OpenCL Best Practices Guide和更短的数组语法。像，

private static boolean repeatedChars(String s) {
    if (s == null) {
        return false;
    }
    char[] chars = s.toCharArray();
    for (int i = 0; i < chars.length - 1; i++) {
        if (chars[i] == chars[i + 1]) {
            return true;
        }
    }
    return false;
}

然后，您可以使用lambda根据没有重复字符的words过滤Scanner sc = new Scanner(System.in); System.out.print("Enter the number or words: "); String[] words = new String[sc.nextInt()]; System.out.print("Enter the strings: "); for (int i = 0; i < words.length; i++) { words[i] = sc.next(); } System.out.println(Arrays.stream(words).filter(s -> !repeatedChars(s)) .collect(Collectors.joining(" ")));并使用String.toCharArray()收集

NONE

，如果您需要显示Predicate<String>消息，可以重新使用Predicate<String> pred = s -> !repeatedChars(s); if (Arrays.stream(words).anyMatch(pred)) { System.out.println(Arrays.stream(words).filter(pred).collect(Collectors.joining(" "))); } else { System.out.println("NONE"); }

junkdf = df = DataFrame({'User_ID': [340,558,558,558,983,422,100,740,740],
'Transaction_Type_x': ['Purchase', 'Purchase', 'Purchase', 'Purchase', 'Purchase', 'Purchase', 'Purchase', 'Purchase', 'Purchase'],
'Rev/Payout_x': [50, 20, 20,28,37,50,40,50,55],
'Cohort_x': ['2010-01','2010-01','2010-02','2010-2','2010-02','2010-02','2010-03','2010-04','2010-04'],
'Transaction_Type_y': ['Sale','Sale','Sale','Sale','Sale','Sale','Sale','Sale','Sale'],
 'Rev/Payout_y': [33,42,66,69,100,22,19,98,39],
 'Cohort_y' : ['2010-03','2010-01','2010-01','2010-04','2010-04','2010-02','2010-01','2010-03','2010-02']})  

junkdf2 =junkdf[['User_ID','Transaction_Type_x','Rev/Payout_x','Cohort_x', 
'Transaction_Type_y','Rev/Payout_y','Cohort_y']]
junkdf2

Answer 2

您的问题有一个简单的解决方法。如果没有任何连续重复，不要立即打印每个单词，而是将其添加到末尾有空格的String变量中，以便每个单词用空格分隔。运行循环后，检查标志是否为false，如果为false则打印NONE。但是，如果为true，则打印结果字符串，在最后添加.trim（）的所有内容。

for (int i = 0; i < words.length; i++) {
    words[i] = sc.next();
}
String result = ""; /*This is the string that holds all the strings that you need to print.*/

for(int i=0;i<words.length;i++){
    int j;
    for(j=1;j<words[i].length();j++) {
        if(words[i].charAt(j) == words[i].charAt(j-1)){
            break;
        }
    }
    if(j==words[i].length()){
        truth = true;
        result  = result + (words[i]+" ");
    }
}
if(!truth){
    System.out.println("NONE");
}
else{
    System.out.println(result.trim()); /*The trim function removes any redundant space in the beginning and the end of the string.*/
}

当然这样做会浪费很多堆内存，但我想这是一个小型的学习项目。但是，请查看StringBuilder如何使用它来避免在堆中创建大量内存！

如何在输出结束时摆脱空间？

2 个答案:

功能使逻辑可读