这里有一个优雅的，非递归的O(n!)解:

public static StringBuilder[] permutations(String s) {
        if (s.length() == 0)
            return null;
        int length = fact(s.length());
        StringBuilder[] sb = new StringBuilder[length];
        for (int i = 0; i < length; i++) {
            sb[i] = new StringBuilder();
        }
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            int times = length / (i + 1);
            for (int j = 0; j < times; j++) {
                for (int k = 0; k < length / times; k++) {
                    sb[j * length / times + k].insert(k, ch);
                }
            }
        }
        return sb;
    }

让我们以输入abc为例。

从集合(["c"])中的最后一个元素(c)开始，然后将最后第二个元素(b)添加到它的前面，末尾和中间的每个可能位置，使其["bc"， "cb"]，然后以同样的方式将后面的下一个元素(a)添加到集合中的每个字符串中，使其:

"a" + "bc" = ["abc", "bac", "bca"]  and  "a" + "cb" = ["acb" ,"cab", "cba"] 

因此整个排列:

["abc", "bac", "bca","acb" ,"cab", "cba"]

代码:

public class Test 
{
    static Set<String> permutations;
    static Set<String> result = new HashSet<String>();

    public static Set<String> permutation(String string) {
        permutations = new HashSet<String>();

        int n = string.length();
        for (int i = n - 1; i >= 0; i--) 
        {
            shuffle(string.charAt(i));
        }
        return permutations;
    }

    private static void shuffle(char c) {
        if (permutations.size() == 0) {
            permutations.add(String.valueOf(c));
        } else {
            Iterator<String> it = permutations.iterator();
            for (int i = 0; i < permutations.size(); i++) {

                String temp1;
                for (; it.hasNext();) {
                    temp1 = it.next();
                    for (int k = 0; k < temp1.length() + 1; k += 1) {
                        StringBuilder sb = new StringBuilder(temp1);

                        sb.insert(k, c);

                        result.add(sb.toString());
                    }
                }
            }
            permutations = result;
            //'result' has to be refreshed so that in next run it doesn't contain stale values.
            result = new HashSet<String>();
        }
    }

    public static void main(String[] args) {
        Set<String> result = permutation("abc");

        System.out.println("\nThere are total of " + result.size() + " permutations:");
        Iterator<String> it = result.iterator();
        while (it.hasNext()) {
            System.out.println(it.next());
        }
    }
}

我的实现基于Mark Byers上面的描述:

    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;
        }
    }

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。

基于Mark Byers的回答，我想出了这个解决方案:

JAVA

public class Main {

    public static void main(String[] args) {
        myPerm("ABCD", 0);
    }

    private static void myPerm(String str, int index)
    {
        if (index == str.length()) System.out.println(str);

        for (int i = index; i < str.length(); i++)
        {
            char prefix = str.charAt(i);
            String suffix = str.substring(0,i) + str.substring(i+1);

            myPerm(prefix + suffix, index + 1);
        }
    }
}

C#

我还使用新的c# 8.0范围操作符在c#中编写了该函数

    class Program
    {
        static void Main(string[] args)
        {
            myPerm("ABCD", 0);
        }

        private static void myPerm(string str, int index)
        {
            if (index == str.Length) Console.WriteLine(str);

            for (int i = index; i < str.Length; i++)
            {
                char prefix = str[i];
                string suffix = str[0..i] + str[(i + 1)..];

                myPerm(prefix + suffix, index + 1);
            }
        }
    

我们只是把每个字母放在开头，然后排列。 第一次迭代是这样的:

/*
myPerm("ABCD",0)  
  prefix = "A"  
  suffix = "BCD"  
  myPerm("ABCD",1)  
    prefix = "B"  
    suffix = "ACD"  
    myPerm("BACD",2)  
      prefix = "C"  
      suffix = "BAD"  
      myPerm("CBAD",3)  
        prefix = "D"  
        suffix = "CBA"  
        myPerm("DCBA",4)  
          Console.WriteLine("DCBA")
*/

这是一个具有O(n!)时间复杂度的算法，具有纯递归和直观。

public class words {
static String combinations;
public static List<String> arrlist=new ArrayList<>();
public static void main(String[] args) {
    words obj = new words();

    String str="premandl";
    obj.getcombination(str, str.length()-1, "");
    System.out.println(arrlist);

}


public void getcombination(String str, int charIndex, String output) {

    if (str.length() == 0) {
        arrlist.add(output);
        return ;
    }

    if (charIndex == -1) {
        return ;
    }

    String character = str.toCharArray()[charIndex] + "";
    getcombination(str, --charIndex, output);

    String remaining = "";

    output = output + character;

    remaining = str.substring(0, charIndex + 1) + str.substring(charIndex + 2);

    getcombination(remaining, remaining.length() - 1, output);

}

}