简单的递归c++实现如下所示:

#include <iostream>

void generatePermutations(std::string &sequence, int index){
    if(index == sequence.size()){
        std::cout << sequence << "\n";
    } else{
        generatePermutations(sequence, index + 1);
        for(int i = index + 1 ; i < sequence.size() ; ++i){
            std::swap(sequence[index], sequence[i]);
            generatePermutations(sequence, index + 1);
            std::swap(sequence[index], sequence[i]);            
        }
    }
}

int main(int argc, char const *argv[])
{
    std::string str = "abc";
    generatePermutations(str, 0);
    return 0;
}

输出:

abc
acb
bac
bca
cba
cab

更新

如果想要存储结果，可以将vector作为函数调用的第三个参数传递。此外，如果您只想要唯一的排列，您可以使用集合。

#include <iostream>
#include <vector>
#include <set>

void generatePermutations(std::string &sequence, int index, std::vector <std::string> &v){
    if(index == sequence.size()){
        //std::cout << sequence << "\n";
        v.push_back(sequence);
    } else{
        generatePermutations(sequence, index + 1, v);
        for(int i = index + 1 ; i < sequence.size() ; ++i){
            std::swap(sequence[index], sequence[i]);
            generatePermutations(sequence, index + 1, v);
            std::swap(sequence[index], sequence[i]);            
        }
    }
}

int main(int argc, char const *argv[])
{
    std::string str = "112";
    std::vector <std::string> permutations;
    generatePermutations(str, 0, permutations);
    std::cout << "Number of permutations " << permutations.size() << "\n";
    for(const std::string &s : permutations){
        std::cout << s << "\n";
    }
    std::set <std::string> uniquePermutations(permutations.begin(), permutations.end());
    std::cout << "Number of unique permutations " << uniquePermutations.size() << "\n";
    for(const std::string &s : uniquePermutations){
        std::cout << s << "\n";
    }
    return 0;
}

输出:

Number of permutations 6
112
121
112
121
211
211
Number of unique permutations 3
112
121
211

//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--]);
        }
    }
}

这是另一个更简单的方法来做一个字符串的排列。

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

}

下面是两个c#版本(仅供参考): 1. 打印所有排列 2. 返回所有排列

算法的基本要点是(可能下面的代码更直观-尽管如此，下面的代码是做什么的一些解释): -从当前索引到集合的其余部分，交换当前索引处的元素 -递归地获得下一个索引中剩余元素的排列 -恢复秩序，通过重新交换

注意:上述递归函数将从起始索引中调用。

private void PrintAllPermutations(int[] a, int index, ref int count)
        {
            if (index == (a.Length - 1))
            {
                count++;
                var s = string.Format("{0}: {1}", count, string.Join(",", a));
                Debug.WriteLine(s);
            }
            for (int i = index; i < a.Length; i++)
            {
                Utilities.swap(ref a[i], ref a[index]);
                this.PrintAllPermutations(a, index + 1, ref count);
                Utilities.swap(ref a[i], ref a[index]);
            }
        }
        private int PrintAllPermutations(int[] a)
        {
            a.ThrowIfNull("a");
            int count = 0;
            this.PrintAllPermutations(a, index:0, count: ref count);
            return count;
        }

版本2(与上面相同-但返回排列而不是打印)

private int[][] GetAllPermutations(int[] a, int index)
        {
            List<int[]> permutations = new List<int[]>();
            if (index == (a.Length - 1))
            {
                permutations.Add(a.ToArray());
            }

            for (int i = index; i < a.Length; i++)
            {
                Utilities.swap(ref a[i], ref a[index]);
                var r = this.GetAllPermutations(a, index + 1);
                permutations.AddRange(r);
                Utilities.swap(ref a[i], ref a[index]);
            }
            return permutations.ToArray();
        }
        private int[][] GetAllPermutations(int[] p)
        {
            p.ThrowIfNull("p");
            return this.GetAllPermutations(p, 0);
        }

单元测试

[TestMethod]
        public void PermutationsTests()
        {
            List<int> input = new List<int>();
            int[] output = { 0, 1, 2, 6, 24, 120 };
            for (int i = 0; i <= 5; i++)
            {
                if (i != 0)
                {
                    input.Add(i);
                }
                Debug.WriteLine("================PrintAllPermutations===================");
                int count = this.PrintAllPermutations(input.ToArray());
                Assert.IsTrue(count == output[i]);
                Debug.WriteLine("=====================GetAllPermutations=================");
                var r = this.GetAllPermutations(input.ToArray());
                Assert.IsTrue(count == r.Length);
                for (int j = 1; j <= r.Length;j++ )
                {
                    string s = string.Format("{0}: {1}", j,
                        string.Join(",", r[j - 1]));
                    Debug.WriteLine(s);
                }
                Debug.WriteLine("No.OfElements: {0}, TotalPerms: {1}", i, count);
            }
        }

下面是一个java实现:

/* All Permutations of a String */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Complexity O(n*n!) */
class Ideone
{
     public static ArrayList<String> strPerm(String str, ArrayList<String> list)
     {
        int len = str.length();
        if(len==1){
            list.add(str);
            return list;
        }

        list = strPerm(str.substring(0,len-1),list);
        int ls = list.size();
        char ap = str.charAt(len-1);
        for(int i=0;i<ls;i++){
            String temp = list.get(i);
            int tl = temp.length();
            for(int j=0;j<=tl;j++){
                list.add(temp.substring(0,j)+ap+temp.substring(j,tl));  
            }
        }

        while(true){
            String temp = list.get(0);
            if(temp.length()<len)
                list.remove(temp);
            else
                break;
        }

        return list;
    }

    public static void main (String[] args) throws java.lang.Exception
    {
        String str = "abc";
        ArrayList<String> list = new ArrayList<>();

        list = strPerm(str,list);
        System.out.println("Total Permutations : "+list.size());
        for(int i=0;i<list.size();i++)
            System.out.println(list.get(i));

    }
}

http://ideone.com/nWPb3k

这就是我通过对排列和递归函数调用的基本理解所做的。虽然要花点时间，但都是独立完成的。

public class LexicographicPermutations {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    String s="abc";
    List<String>combinations=new ArrayList<String>();
    combinations=permutations(s);
    Collections.sort(combinations);
    System.out.println(combinations);
}

private static List<String> permutations(String s) {
    // TODO Auto-generated method stub
    List<String>combinations=new ArrayList<String>();
    if(s.length()==1){
        combinations.add(s);
    }
    else{
        for(int i=0;i<s.length();i++){
            List<String>temp=permutations(s.substring(0, i)+s.substring(i+1));
            for (String string : temp) {
                combinations.add(s.charAt(i)+string);
            }
        }
    }
    return combinations;
}}

生成输出为[abc, acb, bac, bca, cab, cba]。

它背后的基本逻辑是

对于每个字符，将其视为第一个字符，并找出剩余字符的组合。例[abc](abc的组合)->。

a->[bc](a x Combination of (bc))->{abc,acb} b->[ac](b x组合(ac))->{bac,bca} c->[ab](c x Combination of (ab))->{cab,cba}

然后递归地分别调用每个[bc]，[ac]和[ab]。