基于Mark Byers的回答，我的python实现:

def permutations(string):
    if len(string) == 1:
        return [string]
    permutations=[]
    for i in range(len(string)):
        for perm in permutations(string[:i]+string[i+1:]):
            permutations.append(string[i] + perm)
    return permutations

使用Es6的字符串排列

使用reduce()方法

Const排列= STR => { If (str.length <= 2) 返回str.length === 2 ?[str, str[1] + str[0]]: [str]; 返回str .split (") .reduce ( (acc, letter, index) => acc.concat(排列(str。Slice (0, index) + str.slice(index + 1))。Map (val =>字母+ val))， ［］ ）; }; console.log(排列(STR));

使用递归。

依次尝试每个字母作为第一个字母，然后使用递归调用找到剩余字母的所有排列。 基本情况是，当输入是空字符串时，唯一的排列就是空字符串。

//插入每个字符到数组列表中

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) + "  ");
    }
}

//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--]);
        }
    }
}

使用位操作可以很容易地做到这一点。“我们都知道，任何给定的有N个元素的集合有2N个可能的子集。如果我们用一个位来表示子集中的每个元素呢?位可以是0或1，因此我们可以用它来表示对应的元素是否属于这个给定的子集。所以每个位模式代表一个子集。”(复制文本)

private void getPermutation(String str)
        {
            if(str==null)
                return;
            Set<String> StrList = new HashSet<String>();
            StringBuilder strB= new StringBuilder();
            for(int i = 0;i < (1 << str.length()); ++i)
            {
                strB.setLength(0); //clear the StringBuilder
                for(int j = 0;j < str.length() ;++j){
                    if((i & (1 << j))>0){  // to check whether jth bit is set
                        strB.append(str.charAt(j));
                    }
                }
                if(!strB.toString().isEmpty())
                    StrList.add(strB.toString());
            }
            System.out.println(Arrays.toString(StrList.toArray()));
        }