如果有人想生成置换来处理它们，而不是通过void方法打印它们：

static List<int[]> permutations(int n) {
    class Perm {
        private final List<int[]> permutations = new ArrayList<>();
        private void perm(int[] array, int step) {
            if (step == 1) permutations.add(array.clone());
            else for (int i = 0; i < step; i++) {
                perm(array, step - 1);
                int j = (step % 2 == 0) ? i : 0;
                swap(array, step - 1, j);
            }
        }
        private void swap(int[] array, int i, int j) {
            int buffer = array[i];
            array[i] = array[j];
            array[j] = buffer;
        }
    }
    int[] nVector  = new int[n];
    for (int i = 0; i < n; i++) nVector [i] = i;
    Perm perm = new Perm();
    perm.perm(nVector, n);
    return perm.permutations;
}

使用es6的字符串置换
使用reduce（）方法

const permutations = str => {
  if (str.length <= 2) 
  return str.length === 2 ? [str, str[1] + str[0]] : [str];
  return str
    .split('')
    .reduce(
      (acc, letter, index) =>
        acc.concat(permutations(str.slice(0, index) + str.slice(index + 1)).map(val => letter + val)),
      [] 
    );
};
console.log(permutations('STR'));

打印给定字符串的所有排列（考虑重复字符）并仅打印唯一字符的java实现如下所示：

import java.util.Set;
import java.util.HashSet;
public class PrintAllPermutations2
{
    public static void main(String[] args)
    {
        String str = "AAC";
    PrintAllPermutations2 permutation = new PrintAllPermutations2();
    Set<String> uniqueStrings = new HashSet<>();
    permutation.permute("", str, uniqueStrings);
}
void permute(String prefixString, String s, Set<String> set)
{
    int n = s.length();
    if(n == 0)
    {
        if(!set.contains(prefixString))
        {
            System.out.println(prefixString);
            set.add(prefixString);
        }
    }
    else
    {
        for(int i=0; i<n; i++)
        {
            permute(prefixString + s.charAt(i), s.substring(0,i) + s.substring(i+1,n), set);
        }
    }
}
}

下面是另一种更简单的字符串排列方法。

public class Solution4 {
public static void main(String[] args) {
    String  a = "Protijayi";
  per(a, 0);
}
static void per(String a  , int start ) {
      //bse case;
    if(a.length() == start) {System.out.println(a);}
    char[] ca = a.toCharArray();
    //swap 
    for (int i = start; i < ca.length; i++) {
        char t = ca[i];
        ca[i] = ca[start];
        ca[start] = t;
        per(new String(ca),start+1);
    }
}//per
}

import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;
public class hello {
    public static void main(String[] args) throws IOException {
        hello h = new hello();
        h.printcomp();
    }
      int fact=1;
    public void factrec(int a,int k){
        if(a>=k)
        {fact=fact*k;
        k++;
        factrec(a,k);
        }
        else
        {System.out.println("The string  will have "+fact+" permutations");
        }
        }
    public void printcomp(){
        String str;
        int k;
        Scanner in = new Scanner(System.in);
        System.out.println("enter the string whose permutations has to b found");
        str=in.next();
        k=str.length();
        factrec(k,1);
        String[] arr =new String[fact];
        char[] array = str.toCharArray();
        while(p<fact)
        printcomprec(k,array,arr);
            // if incase u need array containing all the permutation use this
            //for(int d=0;d<fact;d++)         
        //System.out.println(arr[d]);
    }
    int y=1;
    int p = 0;
    int g=1;
    int z = 0;
    public void printcomprec(int k,char array[],String arr[]){
        for (int l = 0; l < k; l++) {
            for (int b=0;b<k-1;b++){
            for (int i=1; i<k-g; i++) {
                char temp;
                String stri = "";
                temp = array[i];
                array[i] = array[i + g];
                array[i + g] = temp;
                for (int j = 0; j < k; j++)
                    stri += array[j];
                arr[z] = stri;
                System.out.println(arr[z] + "   " + p++);
                z++;
            }
            }
            char temp;
            temp=array[0];
            array[0]=array[y];
            array[y]=temp;
            if (y >= k-1)
                y=y-(k-1);
            else
                y++;
        }
        if (g >= k-1)
            g=1;
        else
            g++;
    }
}

字符串的排列：

public static void main(String args[]) {
    permu(0,"ABCD");
}
static void permu(int fixed,String s) {
    char[] chr=s.toCharArray();
    if(fixed==s.length())
        System.out.println(s);
    for(int i=fixed;i<s.length();i++) {
        char c=chr[i];
        chr[i]=chr[fixed];
        chr[fixed]=c;
        permu(fixed+1,new String(chr));
    }   
}

我的实现基于mark byers的上述描述：

static Set<String> permutations(String str){
        if (str.isEmpty()){
            return Collections.singleton(str);
        }else{
            Set <String> set = new HashSet<>();
            for (int i=0; i<str.length(); i++)
                for (String s : permutations(str.substring(0, i) + str.substring(i+1)))
                    set.add(str.charAt(i) + s);
            return set;
        }
    }

递归是没有必要的，即使你可以直接计算任何排列，这个解决方案使用泛型来排列任何数组。
这里有一个关于这个算法的好信息。
对于c#开发人员来说，这里是更有用的实现。

public static void main(String[] args) {
    String word = "12345";
    Character[] array = ArrayUtils.toObject(word.toCharArray());
    long[] factorials = Permutation.getFactorials(array.length + 1);
    for (long i = 0; i < factorials[array.length]; i++) {
        Character[] permutation = Permutation.<Character>getPermutation(i, array, factorials);
        printPermutation(permutation);
    }
}
private static void printPermutation(Character[] permutation) {
    for (int i = 0; i < permutation.length; i++) {
        System.out.print(permutation[i]);
    }
    System.out.println();
}

该算法具有o（n）时间和空间复杂度来计算每个置换。

public class Permutation {
    public static <T> T[] getPermutation(long permutationNumber, T[] array, long[] factorials) {
        int[] sequence = generateSequence(permutationNumber, array.length - 1, factorials);
        T[] permutation = generatePermutation(array, sequence);
        return permutation;
    }
    public static <T> T[] generatePermutation(T[] array, int[] sequence) {
        T[] clone = array.clone();
        for (int i = 0; i < clone.length - 1; i++) {
            swap(clone, i, i + sequence[i]);
        }
        return clone;
    }
    private static int[] generateSequence(long permutationNumber, int size, long[] factorials) {
        int[] sequence = new int[size];
        for (int j = 0; j < sequence.length; j++) {
            long factorial = factorials[sequence.length - j];
            sequence[j] = (int) (permutationNumber / factorial);
            permutationNumber = (int) (permutationNumber % factorial);
        }
        return sequence;
    }
    private static <T> void swap(T[] array, int i, int j) {
        T t = array[i];
        array[i] = array[j];
        array[j] = t;
    }
    public static long[] getFactorials(int length) {
        long[] factorials = new long[length];
        long factor = 1;
        for (int i = 0; i < length; i++) {
            factor *= i <= 1 ? 1 : i;
            factorials[i] = factor;
        }
        return factorials;
    }
}

另一种简单的方法是循环字符串，选择尚未使用的字符并将其放入缓冲区，继续循环，直到缓冲区大小等于字符串长度。我更喜欢这种回溯解决方案，因为：
易懂
易于避免重复
输出被排序
以下是java代码：

List<String> permute(String str) {
  if (str == null) {
    return null;
  }
  char[] chars = str.toCharArray();
  boolean[] used = new boolean[chars.length];
  List<String> res = new ArrayList<String>();
  StringBuilder sb = new StringBuilder();
  Arrays.sort(chars);
  helper(chars, used, sb, res);
  return res;
}
void helper(char[] chars, boolean[] used, StringBuilder sb, List<String> res) {
  if (sb.length() == chars.length) {
    res.add(sb.toString());
    return;
  }
  for (int i = 0; i < chars.length; i++) {
    // avoid duplicates
    if (i > 0 && chars[i] == chars[i - 1] && !used[i - 1]) {
      continue;
    }
    // pick the character that has not used yet
    if (!used[i]) {
      used[i] = true;
      sb.append(chars[i]);
      helper(chars, used, sb, res);
      // back tracking
      sb.deleteCharAt(sb.length() - 1);
      used[i] = false;
    }
  }
}

输入str:1231
输出列表：{1123、1132、1213、1231、1312、1321、2113、2131、2311、3112、3121、3211}
请注意，输出已排序，并且没有重复的结果。

//Rotate and create words beginning with all letter possible and push to stack 1
//Read from stack1 and for each word create words with other letters at the next location by rotation and so on 
/*  eg : man
    1. push1 - man, anm, nma
    2. pop1 - nma ,  push2 - nam,nma
       pop1 - anm ,  push2 - amn,anm
       pop1 - man ,  push2 - mna,man
* /
public class StringPermute {
    static String str;
    static String word;
    static int top1 = -1;
    static int top2 = -1;
    static String[] stringArray1;
    static String[] stringArray2;
    static int strlength = 0;
    public static void main(String[] args) throws IOException {
        System.out.println("Enter String : ");
        InputStreamReader isr = new InputStreamReader(System.in);
        BufferedReader bfr = new BufferedReader(isr);
        str = bfr.readLine();
        word = str;
        strlength = str.length();
        int n = 1;
        for (int i = 1; i <= strlength; i++) {
            n = n * i;
        }
        stringArray1 = new String[n];
        stringArray2 = new String[n];
        push(word, 1);
        doPermute();
        display();
    }
    public static void push(String word, int x) {
        if (x == 1)
            stringArray1[++top1] = word;
        else
            stringArray2[++top2] = word;
    }
    public static String pop(int x) {
        if (x == 1)
            return stringArray1[top1--];
        else
            return stringArray2[top2--];
    }
    public static void doPermute() {
        for (int j = strlength; j >= 2; j--)
            popper(j);
    }
    public static void popper(int length) {
        // pop from stack1 , rotate each word n times and push to stack 2
        if (top1 > -1) {
            while (top1 > -1) {
                word = pop(1);
                for (int j = 0; j < length; j++) {
                    rotate(length);
                    push(word, 2);
                }
            }
        }
        // pop from stack2 , rotate each word n times w.r.t position and push to
        // stack 1
        else {
            while (top2 > -1) {
                word = pop(2);
                for (int j = 0; j < length; j++) {
                    rotate(length);
                    push(word, 1);
                }
            }
        }
    }
    public static void rotate(int position) {
        char[] charstring = new char[100];
        for (int j = 0; j < word.length(); j++)
            charstring[j] = word.charAt(j);
        int startpos = strlength - position;
        char temp = charstring[startpos];
        for (int i = startpos; i < strlength - 1; i++) {
            charstring[i] = charstring[i + 1];
        }
        charstring[strlength - 1] = temp;
        word = new String(charstring).trim();
    }
    public static void display() {
        int top;
        if (top1 > -1) {
            while (top1 > -1)
                System.out.println(stringArray1[top1--]);
        } else {
            while (top2 > -1)
                System.out.println(stringArray2[top2--]);
        }
    }
}

//将每个字符插入arraylist

static ArrayList al = new ArrayList();
private static void findPermutation (String str){
    for (int k = 0; k < str.length(); k++) {
        addOneChar(str.charAt(k));
    }
}
//insert one char into ArrayList
private static void addOneChar(char ch){
    String lastPerStr;
    String tempStr;
    ArrayList locAl = new ArrayList();
    for (int i = 0; i < al.size(); i ++ ){
        lastPerStr = al.get(i).toString();
        //System.out.println("lastPerStr: " + lastPerStr);
        for (int j = 0; j <= lastPerStr.length(); j++) {
            tempStr = lastPerStr.substring(0,j) + ch + 
                    lastPerStr.substring(j, lastPerStr.length());
            locAl.add(tempStr);
            //System.out.println("tempStr: " + tempStr);
        }
    }
    if(al.isEmpty()){
        al.add(ch);
    } else {
        al.clear();
        al = locAl;
    }
}
private static void printArrayList(ArrayList al){
    for (int i = 0; i < al.size(); i++) {
        System.out.print(al.get(i) + "  ");
    }
}

无递归的java实现

public Set<String> permutate(String s){
    Queue<String> permutations = new LinkedList<String>();
    Set<String> v = new HashSet<String>();
    permutations.add(s);
    while(permutations.size()!=0){
        String str = permutations.poll();
        if(!v.contains(str)){
            v.add(str);
            for(int i = 0;i<str.length();i++){
                String c = String.valueOf(str.charAt(i));
                permutations.add(str.substring(i+1) + c +  str.substring(0,i));
            }
        }
    }
    return v;
}

我们可以使用阶乘计算以特定字母开头的字符串数量。
示例：以输入为例 d . (3!) == 6 字符串将以字符串的每个字母开头 d .

static public int facts(int x){
    int sum = 1;
    for (int i = 1; i < x; i++) {
        sum *= (i+1);
    }
    return sum;
}
public static void permutation(String str) {
    char[] str2 = str.toCharArray();
    int n = str2.length;
    int permutation = 0;
    if (n == 1) {
        System.out.println(str2[0]);
    } else if (n == 2) {
        System.out.println(str2[0] + "" + str2[1]);
        System.out.println(str2[1] + "" + str2[0]);
    } else {
        for (int i = 0; i < n; i++) {
            if (true) {
                char[] str3 = str.toCharArray();
                char temp = str3[i];
                str3[i] = str3[0];
                str3[0] = temp;
                str2 = str3;
            }
            for (int j = 1, count = 0; count < facts(n-1); j++, count++) {
                if (j != n-1) {
                    char temp1 = str2[j+1];
                    str2[j+1] = str2[j];
                    str2[j] = temp1;
                } else {
                    char temp1 = str2[n-1];
                    str2[n-1] = str2[1];
                    str2[1] = temp1;
                    j = 1;
                } // end of else block
                permutation++;
                System.out.print("permutation " + permutation + " is   -> ");
                for (int k = 0; k < n; k++) {
                    System.out.print(str2[k]);
                } // end of loop k
                System.out.println();
            } // end of loop j
        } // end of loop i
    }
}

下面是一个简单的java最小递归解决方案：

public static ArrayList<String> permutations(String s) {
    ArrayList<String> out = new ArrayList<String>();
    if (s.length() == 1) {
        out.add(s);
        return out;
    }
    char first = s.charAt(0);
    String rest = s.substring(1);
    for (String permutation : permutations(rest)) {
        out.addAll(insertAtAllPositions(first, permutation));
    }
    return out;
}
public static ArrayList<String> insertAtAllPositions(char ch, String s) {
    ArrayList<String> out = new ArrayList<String>();
    for (int i = 0; i <= s.length(); ++i) {
        String inserted = s.substring(0, i) + ch + s.substring(i);
        out.add(inserted);
    }
    return out;
}

/**Returns an array list containing all
 * permutations of the characters in s. */
public static ArrayList<String> permute(String s) {
    ArrayList<String> perms = new ArrayList<>();
    int slen = s.length();
    if (slen > 0) {
        // Add the first character from s to the perms array list.
        perms.add(Character.toString(s.charAt(0)));
        // Repeat for all additional characters in s.
        for (int i = 1;  i < slen;  ++i) {
            // Get the next character from s.
            char c = s.charAt(i);
            // For each of the strings currently in perms do the following:
            int size = perms.size();
            for (int j = 0;  j < size;  ++j) {
                // 1. remove the string
                String p = perms.remove(0);
                int plen = p.length();
                // 2. Add plen + 1 new strings to perms.  Each new string
                //    consists of the removed string with the character c
                //    inserted into it at a unique location.
                for (int k = 0;  k <= plen;  ++k) {
                    perms.add(p.substring(0, k) + c + p.substring(k));
                }
            }
        }
    }
    return perms;
}