lexicographicallyNextString [1004368]

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JavaX fragment (include)

// credits to stackoverflow.com/users/1442960/marcus
static S lexicographicallyNextString(S input) {
  final int lastCharPosition = input.length()-1;
  String inputWithoutLastChar = input.substring(0, lastCharPosition);
  char lastChar = input.charAt(lastCharPosition);
  char incrementedLastChar = (char) (lastChar + 1);
  // Handle int/char overflow.  This wasn't done above.
  if (incrementedLastChar == ((char) 0)) return input+incrementedLastChar;
  return inputWithoutLastChar+incrementedLastChar;
}

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Snippet ID:

#1004368

Snippet name:

lexicographicallyNextString

Eternal ID of this version:

#1004368/1

Text MD5:

b009d1bcfc20da6a4f4aacea582521d7

Author:

stefan

Category:

javax

Type:

JavaX fragment (include)

Public (visible to everyone):

Yes

Archived (hidden from active list):

Created/modified:

2016-08-14 15:53:43

Source code size:

514 bytes / 10 lines

Pitched / IR pitched:

No / No

Views / Downloads:

814 / 799

Referenced in:

[show references]

1	// credits to stackoverflow.com/users/1442960/marcus
2	static S lexicographicallyNextString(S input) {
3	final int lastCharPosition = input.length()-1;
4	String inputWithoutLastChar = input.substring(0, lastCharPosition);
5	char lastChar = input.charAt(lastCharPosition);
6	char incrementedLastChar = (char) (lastChar + 1);
7	// Handle int/char overflow. This wasn't done above.
8	if (incrementedLastChar == ((char) 0)) return input+incrementedLastChar;
9	return inputWithoutLastChar+incrementedLastChar;
10	}