下面是一个简单的Java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;
}

我一直在学习递归思考，第一个打动我的自然解决方案如下。一个更简单的问题是找到一个短一个字母的字符串的排列。我将假设，并相信我的每一根纤维，我的函数可以正确地找到一个字符串的排列，比我目前正在尝试的字符串短一个字母。

Given a string say 'abc', break it into a subproblem of finding permutations of a string one character less which is 'bc'. Once we have permutations of 'bc' we need to know how to combine it with 'a' to get the permutations for 'abc'. This is the core of recursion. Use the solution of a subproblem to solve the current problem. By observation, we can see that inserting 'a' in all the positions of each of the permutations of 'bc' which are 'bc' and 'cb' will give us all the permutations of 'abc'. We have to insert 'a' between adjacent letters and at the front and end of each permutation. For example

我们有bc

“a”+“bc”=“abc”

“b”+“a”+“c”=“bac”

“b”+“a”=“b”

对于'cb'我们有

a + b = acb

“c”+“a”+“b”=“cab”

“cb”+“a”=“cb”

下面的代码片段将说明这一点。下面是该代码片段的工作链接。

def main():
    result = []
    for permutation in ['bc', 'cb']:
        for i in range(len(permutation) + 1):
            result.append(permutation[:i] + 'a' + permutation[i:])
    return result


if __name__ == '__main__':
    print(main())

完整的递归解将是。下面是完整代码的工作链接。

def permutations(s):
    if len(s) == 1 or len(s) == 0:
        return s
    _permutations = []
    for permutation in permutations(s[1:]):
        for i in range(len(permutation) + 1):
            _permutations.append(permutation[:i] + s[0] + permutation[i:])
    return _permutations


def main(s):
    print(permutations(s))


if __name__ == '__main__':
    main('abc')

一个java实现打印给定字符串的所有排列，考虑重复字符，只打印唯一字符，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);
        }
    }
}
}

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));
    }   
}

如果有人想要生成排列来做一些事情，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;

}

改进的代码相同

    static String permutationStr[];
    static int indexStr = 0;

    static int factorial (int i) {
        if (i == 1)
            return 1;
        else
            return i * factorial(i-1);
    }

    public static void permutation(String str) {
        char strArr[] = str.toLowerCase().toCharArray();
        java.util.Arrays.sort(strArr);

        int count = 1, dr = 1;
        for (int i = 0; i < strArr.length-1; i++){
            if ( strArr[i] == strArr[i+1]) {
                count++;
            } else {
                dr *= factorial(count);
                count = 1;
            }       
        }
        dr *= factorial(count);

        count = factorial(strArr.length) / dr;

        permutationStr = new String[count];

        permutation("", str);

        for (String oneStr : permutationStr){
            System.out.println(oneStr);
        }
    }

    private static void permutation(String prefix, String str) {
        int n = str.length();
        if (n == 0) {
            for (int i = 0; i < indexStr; i++){
                if(permutationStr[i].equals(prefix))
                    return;
            }        
            permutationStr[indexStr++] = prefix;
        } else {
            for (int i = 0; i < n; i++) {
                permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i + 1, n));
            }
        }
    }