我编写了一个Java程序来查找字符串中没有Hashmap并设置的重复字符。
以下是该计划，
package practice;
public class Duplicate {
public static void main(String[] args) {
String src= "abcad";
char[] srcChar= src.toLowerCase().toCharArray();
int len=srcChar.length;
int j=0;
boolean flag=false;
char ch;
// System.out.println("Length of the String is "+len1);
// System.out.println("Length of the character array is "+len);
int k=0;
for(int i=0;i<len;i++)
{
// System.out.println("i-----> "+i + " and character is "+srcChar[i]);
for(j=0;j<len;j++)
{
// System.out.println("j-----> "+j + " and character is "+srcChar[j]);
if(srcChar[i]==srcChar[j])
{
k++;
}
}
if(k>1)
{
if(srcChar[i]>1)
{
System.out.println("This character "+srcChar[i]+" has repeated "+k+ " time");
}
else
{
System.out.println("There are no characters repeated in the given string");
}
}
k=0;
}
}
}
这里的输出是：
这个角色a重复了2次 这个角色a重复了2次
在这里，我希望输出像 此字符a已重复2次
即。不重复输出两次。由于字符“a”重复两次，输出也会重复两次。
请帮助我输出一次而不是两次。
谢谢，
答案 0 :(得分：1)
private static void duplicateChar(String str){
char[] arr1 = str.toUpperCase().toCharArray();
int length = str.length();
int count = 1;
String s = "";
char c1 = '\u0000';
for(int i=0;i<length;i++){
count = 1;
for(int j=i+1;j<length;j++){
if(arr1[i] == arr1[j]){
count++;
c1 = arr1[i];
}
if(j == (length-1) && c1 != '\u0000' && !s.contains(String.valueOf(c1))){
s = s+" "+String.valueOf(c1)+" No of times: "+count+"\n";
}
}
}
System.out.println("\nDuplicate char are:\n"+s);
}
答案 1 :(得分：0)
你可以制作一个2维数组，2个宽，源字符串高度。在此数组中，您可以在替换字符时存储该字符，并将其添加到替换它的次数。
类似的东西（我不知道这些计数器是否正确）：
replacements[j][0] = charAt(j);
replacements[j][1] += 1;
您必须检查您要替换的字符是否已存在于此数组中，并且只能打印数组中的元素（如果它们不为空。）
您在原始循环后打印它。
答案 2 :(得分：0)
您需要修复的只是从i而不是0开始第二个循环。
for (int i = 0; i < len; i++) {
for (j = i; j < len; j++) {
...
}
...
}
答案 3 :(得分：0)
<强>进口：
import java.util.ArrayList;
import java.util.List;
<强>代码：
public static void main(String args[]) {
String input = "abcad"; // Input value
char[] chars = input.toLowerCase().toCharArray(); // Creates ArrayList
// of all characters
// in the String
List<Character> charR = new ArrayList<>(); // Creates a List used to
// saving the Characters it
// has saved
List<Integer> valR = new ArrayList<>(); // Creates a List that will
// store how many times a
// character is repeated
for (int i = 0; i < chars.length; i++) { // Loop through items in the
// ArrayList
char c = chars[i]; // Create Character value containing the value of
// the item at the "i" index of the ArrayList
if (charR.contains(c)) { // If the List contains item...
for (int i2 = 0; i2 < charR.size(); i2++) { // Loop through its
// items
if (charR.get(i2).equals(c)) { // If you find a match...
valR.set(i2, valR.get(i2) + 1); // Increase repeated
// value by 1
i2 = charR.size(); // Stop loop
} else { // Else...
i2++; // Increase index by 1
}
}
} else { // Else...
charR.add(c); // Add the Character to the List
valR.add(1); // Add the value 1 to the List (Meaning that the
// Character repeated once)
}
}
for (int i = 0; i < charR.size(); i++) { // Loop through all the items
// in the List
System.out.println("'" + charR.get(i) + "' : " + valR.get(i)); // Display
// what
// the
// character
// is
// and
// how
// many
// times
// it
// was
// repeated
}
}
<强>输出：
'a' : 2
'b' : 1
'c' : 1
'd' : 1
答案 4 :(得分：0)
class PrintDuplicateCharacter
{
public static void main(String[] args)
{
String str = "HelloJava";
char[] ch = str.toCharArray();
int i=0,j=0;
for(i=0;i<ch.length;i++)
{
int count = 0 ;
for( j = i+1;j<ch.length;j++)
{// 4 6 , 8 , 10
if(ch[i] == ch[j] )
{
count++;
}
}
if(count != 0)
{
System.out.print(str.charAt(i) + " Occured " + count + " time");
}
}
}
}
答案 5 :(得分：0)
char[] array=value.toCharArray();
int count=0;
char ch;
for(int i=0;i<array.length-1;i++)
{
ch=array[i];
count=1;
if(ch!='#'){
for(int j=i+1;j<array.length;j++)
{
if(ch==array[j]){
count++;
array[j]='#';
}
}
if(count>1)
{
System.out.println("char is " + ch + "count" + count);
}
}
}
答案 6 :(得分：0)
您还可以使用以下代码来解决此问题：
public static void main（String [] args）{
String src = "abcad";
char[] srcChar = src.toLowerCase().toCharArray();
int len = srcChar.length;
int j = 0;
boolean flag = false;
char ch;
// System.out.println("Length of the String is "+len1);
// System.out.println("Length of the character array is "+len);
int k = 0;
for (int i = 0; i < len; i++) {
// System.out.println("i-----> "+i + " and character is "+srcChar[i]);
for (j = 0 + i; j < len; j++) {
// System.out.println("j-----> "+j + " and character is "+srcChar[j]);
if (srcChar[i] == srcChar[j]) {
k++;
}
}
if (k > 1) {
if (srcChar[i] > 1) {
System.out.println("This character " + srcChar[i] + " has repeated " + k + " time");
} else {
System.out.println("There are no characters repeated in the given string");
}
}
k = 0;
}
}
我们只需要从j = 0 + i开始内部循环；
对于（j = 0 + i; j 这将使您可以观察上面的代码；