java 递归地从字符串中删除重复的字符
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Remove duplicate chars from a String recursively
提问by user647207
I need help to figure how can i remove duplicates chars from a string. It has to be done recursively which is the real problem..
我需要帮助来弄清楚如何从字符串中删除重复的字符。它必须递归完成，这是真正的问题。
public class FEQ2 {
/**
* @param args
*/
public static void removeDups(String s, int firstChar, int secondChar) {
if (s.length() == 1) {
System.out.println(s);
}
char a = s.charAt(firstChar);
if (a == s.charAt(secondChar)) {
s = a + s.substring(secondChar + 1);
}
System.out.println(s);
removeDups(s, firstChar + 1, secondChar + 1);
//return s;
}
public static void main(String[] args) {
//System.out.println(removeDups("AAAABBARRRCC", 1));
removeDups("AAAABBARRRCC", 0 , 1);
}
}
回答by Eng.Fouad
You can do it like this:
你可以这样做：
public static String removeDups(String s)
{
if ( s.length() <= 1 ) return s;
if( s.substring(1,2).equals(s.substring(0,1)) ) return removeDups(s.substring(1));
else return s.substring(0,1) + removeDups(s.substring(1));
}
INPUT: "AAAABBARRRCC"
OUTPUT: "ABARC"
===============
================
EDIT: another way
编辑：另一种方式
public static String removeDups(String s)
{
if ( s.length() <= 1 ) return s;
if( s.substring(1).contains(s.substring(0,1)) ) return removeDups(s.substring(1));
else return s.substring(0,1) + removeDups(s.substring(1));
}
INPUT: "AAAABBARRRCC"
OUTPUT: "BARC"
==============
==============
EDIT: 3rd way
编辑：第三种方式
public static String removeDups(String s)
{
if ( s.length() <= 1 ) return s;
if( s.substring(0,s.length()-1).contains(s.substring(s.length()-1,s.length())) ) return removeDups(s.substring(0,s.length()-1));
else return removeDups(s.substring(0,s.length()-1)) + s.substring(s.length()-1,s.length());
}
INPUT: "AAAABBARRRCC"
OUTPUT: "ABRC"
回答by hugomg
The general trick on doing things recursively is to take all variables and turn them into parameters, and change all assignments into function calls. You might need more than one function for the more complicated stuff, but usually you can turn each loop into a tail-recursivefunction quite easily:
递归做事的一般技巧是取所有变量并将它们转换为参数，并将所有赋值更改为函数调用。对于更复杂的事情，您可能需要多个函数，但通常您可以很容易地将每个循环转换为尾递归函数：
function(){
int i=0; int x=0; //initialize
while(condition){
x = x+i; //update
i = i+1;
}
return x;
}
becomes
变成
function(i,x){ //variables are now parameters
if(condition){
return x;
}else{
return function(i+1, x+i); //update
}
}
main(){
function(0,0); //initialize
===============
================
Here is some duplicate removing code, just for example (it doesn′t do the same thing as yours though)
这是一些重复的删除代码，例如（虽然它和你的不一样）
removeDuplicates(str):
i = str.length-1; out_str = ""; chars_used = []
while(i >= 0):
c = str[i]
if(c not in chars_used):
chars_used.append(c)
out_str += c
i -= 1
return out_str
becomes
变成
remove_duplicates(str, i, out_str, chars_used):
if i < 0:
return out_str
else:
c = str[i]
if c in chars_used:
return remove_duplicates(str, i-1, out_str, chars_used)
else:
return remove_duplicates(str, i-1, out_str+c, chars_used+[c])
回答by James Jithin
Will this help you?
这对你有帮助吗？
public static String getUniqueChars(String realString) {
StringBuilder resultString = null;
try {
List<Character> characterArray = new <Character> ArrayList();
for(char c : realString.toCharArray()) {
characterArray.add(c);
}
resultString = new StringBuilder();
for(Character c : new TreeSet<Character>(characterArray)) {
resultString.append(c.charValue());
}
} catch (Exception e) {
e.printStackTrace();
}
resultString.toString();
}
回答by HITANSU
private static String removeChars(String s) {
int n= s.length();
int i= 0;
int j= 1;
Map<Integer, Boolean> mark= new HashMap<>();
while(j<n) {
if(s.charAt(i)== s.charAt(j)) {
mark.put(i, true);
mark.put(j, true);
if(i== 0) {
i= j+1;
j= i+1;
} else {
i= i-1;
j= j+1;
}
} else {
i= j;
j= j+1;
}
}
String str= "";
for(int k= 0;k<n;k++) {
if(!mark.containsKey(k)) {
str+= s.charAt(k);
}
}
return str;
}
回答by Manny
I think this is simpler:
我认为这更简单：
private static String removeDups(String input){
if (input.length() <= 1) return input;
if (input.charAt(0) == input.charAt(1))
return removeDups(input.substring(1));
else
return input.charAt(0)+removeDups(input.substring(1));
}
回答by Anant Ojha
in C:
在 C：
/* @params:
* src -- input string pointer. String to remove dups from.
* dest -- output string pointer. Passed in as empty string.
* iter -- index from where duplicate removal starts. (0)
* ---------------------------------------------------------
* Purpose:
* Removes duplicate characters from a string recursively.
*/
void remove_Dups_recursively(char *src, char *dest, int iter){
/* base case 1 --> empty or 1 letter string */
if(strlen(src) <= 1){
dest = src;
return;
}
/* base case 2 --> reached end of string */
if(iter == strlen(src)){
return;
}
/* current 'iter' element has been encountered before */
if(strchr(dest, src[iter]) == NULL){
dest[strlen(dest)] = src[iter];
iter++;
remove_Dups_recursively(src, dest, iter);
}
/* current 'iter' element has not been encountered before */
else{
iter++;
remove_Dups_recursively(src, dest, iter);
}
}
/* EXAMPLE: AAABBABCCABA
* OUTPUT: ABC */
