Java 删除字符串中的重复字符
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Remove repeated characters in a string
提问by abedzantout
I need to write a static method that takes a Stringas a parameter and returns a new Stringobtained by replacing every instance of repeated adjacent letters with a single instance of that letter without using regular expressions. For example if I enter "maaaakkee" as a String, it returns "make".
I already tried the following code, but it doesn't seem to display the last character.
Here's my code:
我需要编写一个静态方法，该方法将 aString作为参数并返回一个新方法，该方法通过在String不使用正则表达式的情况下用该字母的单个实例替换重复相邻字母的每个实例而获得。例如，如果我输入“maaaakkee”作为 a String，它会返回“make”。我已经尝试了以下代码，但它似乎没有显示最后一个字符。这是我的代码：
import java.util.Scanner;
public class undouble {
public static void main(String [] args){
Scanner console = new Scanner(System.in);
System.out.println("enter String: ");
String str = console.nextLine();
System.out.println(removeSpaces(str));
}
public static String removeSpaces(String str){
String ourString="";
int j = 0;
for (int i=0; i<str.length()-1 ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}
return ourString;
}
}
采纳答案by Rupesh
The problem is with your condition. You say compare i and i+1 in each iteration and in last iteration you have both i and j pointing to same location so it will never print the last character. Try this unleass you want to use regex to achive this:
问题出在你的病情上。你说在每次迭代中比较 i 和 i+1，在最后一次迭代中你有 i 和 j 指向相同的位置，所以它永远不会打印最后一个字符。试试这个，除非你想使用正则表达式来实现这个：
EDIT:
编辑：
public void removeSpaces(String str){
String ourString="";
for (int i=0; i<str.length()-1 ; i++){
if(i==0){
ourString = ""+str.charAt(i);
}else{
if(str.charAt(i-1) != str.charAt(i)){
ourString = ourString +str.charAt(i);
}
}
}
System.out.println(ourString);
}
回答by Rich
maybe:
也许：
for (int i=1; i<str.length() ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}
回答by Mena
You could use regular expressions for that.
您可以为此使用正则表达式。
For instance:
例如：
String input = "ddooooonnneeeeee";
System.out.println(input.replaceAll("(.)\1{1,}", ""));
Output:
输出：
done
Pattern explanation:
图案说明：
"(.)\\1{1,}"means any character (added to group 1) followed by itself at least once"$1"references contents of group 1
"(.)\\1{1,}"表示任何字符（添加到第 1 组）后跟它自己至少一次"$1"参考第 1 组的内容
回答by kunwar.sangram
More fun with java 7:
使用 Java 7 更有趣：
System.out.println("11223344445555".replaceAll("(?<nums>.+)\k<nums>+","${nums}"));
No more cryptic numbers in regexes.
正则表达式中不再有神秘数字。
回答by GuillaumeAgis
if you cannot use replace or replaceAll, here is an alternative. O(2n), O(N) for stockage and O(N) for creating the string. It removes all repeated chars in the string put them in a stringbuilder.
如果您不能使用 replace 或 replaceAll，这里有一个替代方法。O(2n)、O(N) 用于库存和 O(N) 用于创建字符串。它删除字符串中所有重复的字符，将它们放入字符串生成器中。
input : abcdef , output : abcdef
输入：abcdef，输出：abcdef
input : aabbcdeef, output : cdf
输入：aabbcdeef，输出：cdf
private static String remove_repeated_char(String str)
{
StringBuilder result = new StringBuilder();
HashMap<Character, Integer> items = new HashMap<>();
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == null)
items.put(current, 1);
else
items.put(current, ocurrence + 1);
}
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == 1)
result.append(current);
}
return result.toString();
}
回答by myinterviewexp.com
public static String removeDuplicates(String str) {
公共静态字符串 removeDuplicates(String str) {
String str2 = "" + str.charAt(0);
for (int i = 1; i < str.length(); i++) {
if (str.charAt(i - 1) == str.charAt(i) && i != 0) {
continue;
}
str2 = str2 + str.charAt(i);
}
return str2;
}
回答by user6885473
import java.util.*;
public class string2 {
public static void main(String[] args) {
//removes repeat character from array
Scanner sc=new Scanner(System.in);
StringBuffer sf=new StringBuffer();
System.out.println("enter a string");
sf.append(sc.nextLine());
System.out.println("string="+sf);
int i=0;
while( i<sf.length())
{
int j=1+i;
while(j<sf.length())
{
if(sf.charAt(i)==sf.charAt(j))
{
sf.deleteCharAt(j);
}
else
{
j=j+1;
}
}
i=i+1;
}
System.out.println("string="+sf);
}
}
回答by CodeNinja
Input AABBBccDDD, Output BD Input ABBCDDA, Outout C
输入 AABBccDDD，输出 BD 输入 ABBCDDA，输出 C
private String reducedString(String s){
char[] arr = s.toCharArray();
String newString = "";
Map<Character,Integer> map = new HashMap<Character,Integer>();
map.put(arr[0],1);
for(int index=1;index<s.length();index++)
{
Character key = arr[index];
int value;
if(map.get(key) ==null)
{
value =0;
}
else
{
value = map.get(key);
}
value = value+1;
map.put(key,value);
}
Set<Character> keyset = map.keySet();
for(Character c: keyset)
{
int value = map.get(c);
if(value%2 !=0)
{
newString+=c;
}
}
newString = newString.equals("")?"Empty String":newString;
return newString;
}
回答by Neeraj Gupta
public class RemoveDuplicateCharecterInString {
static String input = new String("abbbbbbbbbbbbbbbbccccd");
static String output = "";
public static void main(String[] args)
{
// TODO Auto-generated method stub
for (int i = 0; i < input.length(); i++) {
char temp = input.charAt(i);
boolean check = false;
for (int j = 0; j < output.length(); j++) {
if (output.charAt(j) == input.charAt(i)) {
check = true;
}
}
if (!check) {
output = output + input.charAt(i);
}
}
System.out.println(" " + output);
}
}
Answer : abcd
答案：abcd
回答by Thu Vo
public class RepeatedChar {
public static void main(String[] args) {
String rS = "maaaakkee";
String outCome= rS.charAt(0)+"";
int count =0;
char [] cA =rS.toCharArray();
for(int i =0; i+1<cA.length; ++i) {
if(rS.charAt(i) != rS.charAt(i+1)) {
outCome += rS.charAt(i+1);
}
}
System.out.println(outCome);
}
}
