Java method to count the number of characters and reverse order different characters
- 2020-04-01 02:02:54
- OfStack
import java.util.ArrayList;
import java.util.List;
public class Test2 {
public static void main(String[] args) {
String src = "A B C D E B C";
//Replace the space
src = src.replaceAll(" ", "") ;
System.out.println(" String after removing Spaces: " + src) ;
List<Character> list = new ArrayList<Character>() ;
int[] bb = new int[256];
char[] cs = src.toCharArray();
//Reverse order
int mid = cs.length / 2 ;
int idx = cs.length -1 ;
for (int i = 0; i < mid ; i++){
char tmp = cs[i] ;
cs[i] = cs[idx] ;
cs[idx] = tmp ;
idx-- ;
}
//Count and filter the same
for (char c : cs) {
if (bb[c] <1) {
list.add(c) ;
}
bb[c] = bb[c] + 1;
}
System.out.println();
for (int i = 0; i < list.size(); i++){
System.out.print(list.get(i)) ;
}
System.out.println() ;
for (int i = 0; i < list.size(); i++){
char c = list.get(i) ;
System.out.println(c + " " + bb[c] + " time ") ;
}
}
}
String a = "abcd, efg";
String b = "(* & ^ % $# @! [] {},. / /; : '? < >" ;
The requirement is to determine whether any character in String a appears in String b, the more efficient the better
* finds whether certain characters appear in another string
*
* @author Java people (java2000.net)
*/
public class Test {
public static void main(String[] args) {
String a = "abcd,efg";
String b = ")(*&^%$#@![]{},.///;:'? <>";
byte[] bb = new byte[256];
char[] cs = b.toCharArray();
for (char c : cs) {
bb[c] = 1;
}
cs = a.toCharArray();
for (char c : cs) {
if (bb[c] == 1) {
System.out.println(c);
}
}
}
}