替换字符串的第n个字符但忽略空格的最佳方法？

Question

I'm trying to figure out the best way to replace the nth character of a string but ignore the space when looping. For example, if I was to change every 5th character of the String mary had a little lamb to z , it should return mary zad azlittze lazb

One way I thought would be to remove all space ( maryhadalittlelamb ), then change all the 5th character to z ( maryzadazlittzelazb ) and then reference the original string, find the index of " " and insert it into maryzadazlittzelazb

But this doesn't seem very elegant and I'm sure theres a better way to do this, could someone please advise?

Thanks!

Answer 1

I would use String.toCharArray() , then iterate with a regular for loop and test if each character is not whitespace with Character.isWhitespace(char) . If it's not, increment the second counter (here named p ) and check if that value is divisible by five. If so, set it to z . Finally, create a new String based on the edited char[] . Like,

String str = "mary had a little lamb";
char[] arr = str.toCharArray();
for (int i = 0, p = 0; i < arr.length; i++) {
    if (!Character.isWhitespace(arr[i]) && ++p % 5 == 0) {
        arr[i] = 'z';
    }
}
System.out.println(new String(arr));

I get (as I mentioned in the comments)

mary zad a lzttle zamb

Also, because it might not be very clear, the complex if above is equivalent to

if (!Character.isWhitespace(arr[i])) {
    p++;
    if (p % 5 == 0) {
        arr[i] = 'z';
    }
}

Answer 2

For completeness, my suggestion would have looked like this:

public static void main(String... args) {
  Pattern p = Pattern.compile("((\\S\\s*){4})\\S");
  Matcher m = p.matcher("mary had a little lamb");
  StringBuffer sb = new StringBuffer();
  while (m.find()) {
    m.appendReplacement(sb, "$1z");
  }
  m.appendTail(sb);
  System.out.println(sb.toString()); 
}