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Determine if a string has all unique characters: Difference between revisions
Determine if a string has all unique characters (view source)
Revision as of 20:26, 25 February 2020
, 4 years ago→{{header|Haskell}}
(Added C#) |
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Line 1,077:
"XYZ ZYX" (7) -> 'X' (0x58) at 0, 6
"1234567890ABCDEFGHIJKLMN0PQRSTUVWXYZ" (36) -> '0' (0x30) at 9, 24</pre>
=={{header|Java}}==
<lang java>
import java.util.HashMap;
import java.util.Map;
// Title: Determine if a string has all unique characters
public class StringUniqueCharacters {
public static void main(String[] args) {
System.out.printf("%-40s %2s %10s %8s %s %s%n", "String", "Length", "All Unique", "1st Diff", "Hex", "Positions");
System.out.printf("%-40s %2s %10s %8s %s %s%n", "------------------------", "------", "----------", "--------", "---", "---------");
for ( String s : new String[] {"", ".", "abcABC", "XYZ ZYX", "1234567890ABCDEFGHIJKLMN0PQRSTUVWXYZ"} ) {
processString(s);
}
}
private static void processString(String input) {
Map<Character,Integer> charMap = new HashMap<>();
char dup = 0;
int index = 0;
int pos1 = -1;
int pos2 = -1;
for ( char key : input.toCharArray() ) {
index++;
if ( charMap.containsKey(key) ) {
dup = key;
pos1 = charMap.get(key);
pos2 = index;
break;
}
charMap.put(key, index);
}
String unique = dup == 0 ? "yes" : "no";
String diff = dup == 0 ? "" : "'" + dup + "'";
String hex = dup == 0 ? "" : Integer.toHexString(dup).toUpperCase();
String position = dup == 0 ? "" : pos1 + " " + pos2;
System.out.printf("%-40s %-6d %-10s %-8s %-3s %-5s%n", input, input.length(), unique, diff, hex, position);
}
}
</lang>
{{Out}}
<pre>
String Length All Unique 1st Diff Hex Positions
------------------------ ------ ---------- -------- --- ---------
0 yes
. 1 yes
abcABC 6 yes
XYZ ZYX 7 no 'Z' 5A 3 5
1234567890ABCDEFGHIJKLMN0PQRSTUVWXYZ 36 no '0' 30 10 25
</pre>
=={{header|JavaScript}}==
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