在java中比较字符串和打印剩余字符时出现意外输出？

Question

这里有两条线

String first = "stack overflow";
String second = "java programming"; 

因此，我需要输出为：

stck eflow j prgmming

即，从第一个字符串中提取字符，并与第二个字符串中剩余的所有字符进行比较，如果存在，则将其删除。下面是我的代码，给出了一些意想不到的结果：

String first = "stack overflow";
String second = "java programming";
for (int i = 0; i < first.length(); i++) {
    out.println("char position :" + first.charAt(i));
    for (int j = 0; j < second.length(); j++) {                    
        if (first.charAt(i) == second.charAt(j)) {
            first = first.substring(0, i) + first.substring(i + 1);
            second = second.substring(0, j) + second.substring(j + 1);
        }
    }
}
out.println(first+second);

Answer 1

可以将StringBuffer与字符串一起使用。从StringBuffer对象而不是String对象中删除必要的字符

    String first = "stack overflow";
    String second = "java programming";
    StringBuffer firstBuffer = new StringBuffer(first);
    StringBuffer secondBuffer = new StringBuffer(second);
    for (int i = 0; i < first.length(); i++) {
        out.println("char position :" + first.charAt(i));
        for (int j = 0; j < second.length(); j++) {                    
            if (first.charAt(i) == second.charAt(j)) {
                firstBuffer.deleteCharAt(i);
                secondBuffer.deleteCharAt(j);
            }
        }
    }
    System.out.println(firstBuffer.toString()+secondBuffer.toString());

Answer 2

我会先看每个单词中的字母，然后再打印声明。既然你说的要求是打印它，而不是返回一个值，那就是我要做的。如果这是不正确的，你可以构建字符串，就像我打印并返回字符串一样。鉴于我们的评论讨论，我将尝试让这更像是一个教程，而不是一个答案

该准则做出以下假设： 两个字符串之间只有空格是有效字符。 两个词都在同一个格里

//Getting the letters up front could allow us to improve 
//the efficiency of this algorithm to O(n) from your O(n^2) solution
private Set<Character> getLetters(String word) {
   HashSet<Character> letters= new HashSet<Character>();
   for (char c: word.toCharArray()) {
      //your output prints spaces so I'll skip them
      if (c != ' ')
         letters.add(c);
   }
   return letters;
}

//Breaking out duplicate code into methods is good for code reuse
//This method will print any character not in the Set provided
private void printUnmatchedCharacters(String word, Set<Character> letters) {
   for (char c: word.toCharArray()) {
       //Since this is actually a HashSet contains is fast at O(1)
       if(!letters.contains(c))
          System.out.print(c);
   }
}

public void printResults(String first, String second) {
   printUnmatchedCharacters(first, getLetters(second));
   //You have a space between your words
   System.out.print(' ');
   printUnmatchedCharacters(second, getLetters(first));
}

虽然上面的代码更长，但效率更高，可以分解成更小的代码块。如果出现问题，调试会更容易一些，您可以更容易地进行单元测试。因为你是编程新手，我希望这个答案能帮助你学到一些其他的东西，除了你想要的问题