添加点到system.out.println

Question

基本上，我基本上有一条线，它有一个尖顶在40, 如果添加2个单词是每个10个字母，每个长度留下20个空格 thouse 20个空格应该是充满点......例如 你好................................你

或

堆栈溢出...........................

现在我的代码工作正常，我只是不关于如何在system.out.println中取悦这个想法，以便打印结果t控制台。

我想在使用whileloop打印一次一个点之后，在测试一次打印多少个点后应该有一些打扰的词被输入。

目前，我有这个这显然行不通

{ 
     System.out.println (word1 + ".." + word2); 

     while (wordlegnth1 + wordlegnth2 < LINELENGTH) 

     { 
      System.out.println (word1 + "." + word2); 

      wordlegnth1++; 

     } 

Answer 1

final int LINE_LENGTH = 40; 

String word1 = ...; 
String word2 = ...; 

StringBuilder sb = new StringBuilder(LINE_LENGTH); 
sb.append(word1); 
for (int i = 0; i + word1.length() + word2.length() < LINE_LENGTH; i++) { 
    sb.append("."); 
} 
sb.append(word2); 

System.out.println(sb.toString()); 

注意：在使用StringBuilder的是为了避免性能下降，由于字符串不变性

Answer 2

/** 
* Formats a given string to 40 characters by prefixing it with word1 and 
* suffixing it with word2 and using dots in the middle to make length of final 
* string = 40. 
* If string will not fit in 40 characters, -1 is returned, otherwise 0 is 
* returned. 
* 
* @param word1 the first word 
* @param word2 the second word 
* @return 0 if string will fit in 40 characters, -1 otherwise 
*/ 
public static int formatString(String word1, String word2) { 
    final int LINELENGTH = 40; 

    // determine how many dots we need 
    int diff = LINELENGTH - (word1.length() + word2.length()); 

    if (diff < 0) { 
     // it's too big 
     System.out.println("string is too big to fit"); 
     return -1; 
    } 
    // add the dots 
    StringBuilder dots = new StringBuilder(); 
    for (int i = 0; i < diff; ++i) { 
     dots.append("."); 
    } 

    // build the final result 
    String result = word1 + dots.toString() + word2; 

    System.out.println(result); 

    return 0; 

} 

public static void main(String[] args) throws Throwable { 
    formatString("stack", "overflow"); 
    String str = ""; 
    for (int i = 0; i < 21; ++i) { 
     str += "a"; 
    } 
    formatString(str, str); 
} 

输出：

stack...........................overflow 
string is too big to fit 

Answer 3

为什么不喜欢的东西开始：

int lengthA = word1.length(); 
int lengthB = word2.length(); 
int required = LINE_LENGHT - (lengthA + lengthB); 

for(int i = 0; i < required; i++) 
{ 
    System.out.print("."); 
} 

它可以改进（尝试字符串非常大例如，或使用StringBuilder而不是System.out.print）。

Answer 4

Apache的公地郎添加到您的类路径和使用StringUtils.leftPad()

System.out.println(str1 + StringUtils.leftPad(str2, 40 - str1.length(), '.')); 

为什么要写做一些别人已经写+测试的方法？