Question

我正在开发一个 Anagram 程序，我正在研究一个名为diff的方法，该方法应该返回一个包含第一个{{1}中的字符的StringBuffer但不是在第二个。因此，例如，如果StringBuffers是StringBuffer和abba，那么我的diff方法应该返回bb。到目前为止，我目前有一个带有if语句的循环，但它不起作用。任何帮助，将不胜感激。谢谢

acca

Answer 1

我认为您正在尝试使用循环在sBuffer1中按一个字符检查一个字符。但是String let1 = String.valueOf(sBuffer1);会为您提供整个sBuffer1字符串。

您需要String let1 = sBuffer1.substring(i, i + 1)从sBuffer1中获取一个字符来检查它是否存在于sBuffer2中。

例如：

public static StringBuffer diff(StringBuffer sBuffer1, StringBuffer sBuffer2) {
    StringBuffer diffVal = new StringBuffer();

    for (int i = 0; i < sBuffer1.length(); i++) {
        String let1 = sBuffer1.substring(i, i + 1); // get the character from sBuffer1
        if (sBuffer2.indexOf(let1) == -1) {
            diffVal.append(let1); // append the character to the diff
        }
    }

    return diffVal;
}

Answer 2

确定这可能有效，你的逻辑有点不对，这段代码很简单。如果第二个字符串缓冲区中不存在该字符，请将其添加到结果SF中。

public StringBuffer diff(){
        StringBuffer diffVal = new StringBuffer();//initialize before you use it.

        for (int i =0; i < sBuffer1.length(); i++){
            String let1 = String.valueOf(sBuffer1.charAt(i))//get the character at the ith position.
            if (sBuffer2.indexOf(let1) == -1 ){
              diffVal.append(let1);
            }

        }
        return diffVal;
    }

Answer 3

试试这个。

   StringBuilder diffVal= new StringBuilder();

            StringBuffer sBuffer1 = new StringBuffer("abbad");//input 1
            StringBuffer sBuffer2 = new StringBuffer("acca");//input 2, you can ignore if you have already passed/defined these
            for (int i =0; i < sBuffer1.length(); i++){             
               if(i >= sBuffer2.length()){//handles difference in input string length
                   diffVal.append(sBuffer1.substring(i, sBuffer1.length()));
                   break;
               }
            if (sBuffer1.charAt(i) != sBuffer2.charAt(i)) {
                diffVal.append(sBuffer1.charAt(i));
            }

            }
           System.out.println(diffVal);// I am printing it here

输出结果为：bbd

如果您在此处使用的字符串不需要StringBuilder

，则建议使用synchronized

Java在StringBuffer中查找字符之间的差异

3 个答案: