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);

}

}

以下是我在《破解编程面试》(P54)一书中提出的解决方案:

/**
 * List permutations of a string.
 * 
 * @param s the input string
 * @return  the list of permutations
 */
public static ArrayList<String> permutation(String s) {
    // The result
    ArrayList<String> res = new ArrayList<String>();
    // If input string's length is 1, return {s}
    if (s.length() == 1) {
        res.add(s);
    } else if (s.length() > 1) {
        int lastIndex = s.length() - 1;
        // Find out the last character
        String last = s.substring(lastIndex);
        // Rest of the string
        String rest = s.substring(0, lastIndex);
        // Perform permutation on the rest string and
        // merge with the last character
        res = merge(permutation(rest), last);
    }
    return res;
}

/**
 * @param list a result of permutation, e.g. {"ab", "ba"}
 * @param c    the last character
 * @return     a merged new list, e.g. {"cab", "acb" ... }
 */
public static ArrayList<String> merge(ArrayList<String> list, String c) {
    ArrayList<String> res = new ArrayList<>();
    // Loop through all the string in the list
    for (String s : list) {
        // For each string, insert the last character to all possible positions
        // and add them to the new list
        for (int i = 0; i <= s.length(); ++i) {
            String ps = new StringBuffer(s).insert(i, c).toString();
            res.add(ps);
        }
    }
    return res;
}

字符串"abcd"的运行输出:

第一步:合并[a]和b: [ba, ab] 步骤2:Merge [ba, ab]和c: [cba, bca, bac, cab, acb, abc] 第三步:Merge [cba, bca, bac, cab, acb, abc]和d: [dcba, cdba, cbad, cbca, bdcad

递归是不必要的，甚至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。

这里有关于这个algorihtm的很好的信息。

对于c#开发人员来说，这里有更有用的实现。

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();
}

该算法计算每个排列的时间和空间复杂度为O(N)。

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;
    }
}

在python中

def perms(in_str, prefix=""):
if not len(in_str) :
    print(prefix)
else:        
    for i in range(0, len(in_str)):
        perms(in_str[:i] + in_str[i + 1:], prefix + in_str[i])

perms('ASD')

基于Heap算法的我的实现:

import java.util.ArrayList;
import java.util.List;

public class PermutationString {
public static List<String> permute(char[] str, int n) {
    List<String> permutations = new ArrayList<>();
    if (n == 1) {
        permutations.add(new String(str));
    }
    else {
        for (int i = 0; i < n; i++) {
            permutations.addAll(permute(str, n-1));
            if (n % 2 == 0) {
                swap(str, i, n-1);
            }
            else {
                swap(str, 0, n-1);
            }
        }
    }
    return permutations;
}


public static void swap(char[] str, int i, int j) {
    char temp = str[i];
    str[i] = str[j];
    str[j] = temp;
}

public static void main(String[] args) {

    List<String> permutations = permute("abcdefgh".toCharArray(), 8);

    System.out.println(permutations);

}
}

时间复杂度为O(n!* n)， O(n)为空间复杂度。

public static void permutation(String str) { 
    permutation("", str); 
}

private static void permutation(String prefix, String str) {
    int n = str.length();
    if (n == 0) System.out.println(prefix);
    else {
        for (int i = 0; i < n; i++)
            permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n));
    }
}

(通过Java编程入门)