使用Java检查字符串中的邻居 [英] Check Neighbours in a String with Java

问题描述

使用字符串(如电话" )，我想知道字符 的邻居>'o'在这种情况下，我尝试使用String迭代器，但这给了我一个在使用charAt()之前或之后，我将超出-1或无限循环范围

in a string like "phone" i want to know the neighbour of the character 'o' in this case 'h' and 'n' i tryed with a String Iterator but that gives me either before or after and with charAt() i will be out of range by -1 or endless loop

String s = textArea.getText();
    
    for( int i = 0; i < s.length(); i++) {
        char ch = s.charAt(i);
        char tz = s.charAt(i--);
                            
            System.out.print(ch);
            if(ch == 'n') {
            System.out.print(tz);
            break;
            }
        }

推荐答案

您可以尝试类似的方法.当然，您仍然可以对代码进行更改以获得预期的结果.

you could try something like this. Of course, you can still do changes to the code to get the expected result.

public void stringSplit(String text)
    {
        char[] letters = text.toCharArray();
        for (int i=0; i<letters.length;i++)
        {
            if (i!=0 && i!=letters.length-1){
                System.out.print("neighbour left: " + letters[i - 1] +'\n');
                System.out.print("actual: " + letters[i]+'\n');
                if (letters[i - 1] == 'n') { //here I used the condition from your code
                    System.out.print("neighbour right: " + letters[i + 1] +'\n');
                    break;
                }
            }
            else if(i==0)
            {
                System.out.print("actual: " + letters[i]+'\n');
                System.out.print("neighbour right: " + letters[i + 1] +'\n');
            }
            else{
                System.out.print("neighbour left: " + letters[i - 1] +'\n');
                System.out.print("actual: " + letters[i]+'\n');

                System.out.println("end of string");
            }
        }
    }

在此版本中，您已检查了所有极端情况.你还有问题吗?

In this version you have all the corner cases checked. Do you still have questions?

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