将点添加到 system.out.println 中

Question

基本上我有一条线的结尾是 40， 如果添加 2 个单词，每个单词长 10 个字母，那么我就剩下 20 个空格 20 个空格应该填满点……例如 你好......................你

或

堆栈...................... ......溢出

现在我的代码工作正常，我只是不知道如何在 system.out.println 中满足这一点，以便将结果打印到控制台。

在测试输入干扰词后应该有多少个点之后，我正在考虑使用 whileloop 一次打印一个点。

目前我有这个显然不起作用

{
       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：使用StringBuilder是为了避免由于String不可变性而导致的性能损失

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

为什么不从以下内容开始：

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（例如尝试非常大的字符串，或者使用 StringBuilder 而不是 System.out.print）。

Answer 4

将 apache 的 commons-lang 添加到您的类路径并使用 StringUtils.leftPad()

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

为什么要编写一个方法来执行其他人已经编写并测试过的操作？