Check Neighbours in a String with Java

问题

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;
            }
        }

回答1:

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");
            }
        }
    }

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

回答2:

s.charAt(i--) subtracts one in loop. It it leads to endless loop. You can try s.charAt(i-1) instead

回答3:

try this :

public static void main(String[] args) {
    String s = "phone";
    char[] arr = neighbour(s, 'o');
    System.out.println("previous : "+arr[0]);
    System.out.println("next : "+arr[1]);
}

public static char[] neighbour(String s , char c){
    int index = s.indexOf(c);
    if(index == -1) throw new Error("character '"+c+"' not exist in \""+s+"\" ");
    char[] arr = new char[2];
    if(index > 0)
        arr[0] = s.charAt(index-1);
    if(index+1 < s.length())
        arr[1] = s.charAt(index+1);
    return arr;
}

this will get the next and previous character off specific char and put them inside array of 2 character.

来源：https://stackoverflow.com/questions/65431509/check-neighbours-in-a-string-with-java

标签

java

loops

character

charat