倒计时Quickperm算法的通用实现，表示#1(可伸缩，非递归)。

/**
 * Generate permutations based on the
 * Countdown <a href="http://quickperm.org/">Quickperm algorithm</>.
 */
public static <T> List<List<T>> generatePermutations(List<T> list) {
    List<T> in = new ArrayList<>(list);
    List<List<T>> out = new ArrayList<>(factorial(list.size()));

    int n = list.size();
    int[] p = new int[n +1];
    for (int i = 0; i < p.length; i ++) {
        p[i] = i;
    }
    int i = 0;
    while (i < n) {
        p[i]--;
        int j = 0;
        if (i % 2 != 0) { // odd?
            j = p[i];
        }
        // swap
        T iTmp = in.get(i);
        in.set(i, in.get(j));
        in.set(j, iTmp);

        i = 1;
        while (p[i] == 0){
            p[i] = i;
            i++;
        }
        out.add(new ArrayList<>(in));
    }
    return out;
}

private static int factorial(int num) {
    int count = num;
    while (num != 1) {
        count *= --num;
    }
    return count;
}

它需要list，因为泛型不能很好地使用数组。

使用Set操作建模“依赖于其他选择的选择”更容易理解相关排列 使用依赖排列，可用的选择减少，因为位置被从左到右的选定字符填充。递归调用的终端条件是测试可用选择集是否为空。当满足终端条件时，置换完成，并存储到“结果”列表中。

public static List<String> stringPermutation(String s) {
    List<String> results = new ArrayList<>();
    Set<Character> charSet = s.chars().mapToObj(m -> (char) m).collect(Collectors.toSet());
    stringPermutation(charSet, "", results);
    return results;
}

private static void stringPermutation(Set<Character> charSet, 
        String prefix, List<String> results) {
    if (charSet.isEmpty()) {
        results.add(prefix);
        return;
    }
    for (Character c : charSet) {
        Set<Character> newSet = new HashSet<>(charSet);
        newSet.remove(c);
        stringPermutation(newSet, prefix + c, results);
    }
} 

该代码可以泛化为一组对象查找排列。在本例中，我使用了一组颜色。

public enum Color{
    ORANGE,RED,BULE,GREEN,YELLOW;
}

public static List<List<Color>> colorPermutation(Set<Color> colors) {
    List<List<Color>> results = new ArrayList<>();
    List<Color> prefix = new ArrayList<>();
    permutation(colors, prefix, results);
    return results;
}

private static <T> void permutation(Set<T> set, List<T> prefix, List<List<T>> results) {
    if (set.isEmpty()) {
        results.add(prefix);
        return;
    }
    for (T t : set) {
        Set<T> newSet = new HashSet<>(set);
        List<T> newPrefix = new ArrayList<>(prefix);
        newSet.remove(t);
        newPrefix.add(t);
        permutation(newSet, newPrefix, results);
    }
} 

测试代码。

public static void main(String[] args) {
    List<String> stringPerm = stringPermutation("abcde");
    System.out.println("# of permutations:" + stringPerm.size());
    stringPerm.stream().forEach(e -> System.out.println(e));

    Set<Color> colorSet = Arrays.stream(Color.values()).collect(Collectors.toSet());
    List<List<Color>> colorPerm = colorPermutation(colorSet);
    System.out.println("# of permutations:" + colorPerm.size());
    colorPerm.stream().forEach(e -> System.out.println(e));
}

import java.io.*;
public class Anagram {

public static void main(String[] args) {
      java.util.Scanner sc=new java.util.Scanner(System.in);
            PrintWriter p=new PrintWriter(System.out,true);
            p.println("Enter Word");
            String a[],s="",st;boolean flag=true;
            int in[],n,nf=1,i,j=0,k,m=0;
            char l[];
            st=sc.next();
            p.println("Anagrams");
            p.println("1 . "+st);
            l=st.toCharArray();
            n=st.length();
            for(i=1;i<=n;i++){
                nf*=i;
            }

            i=1;
            a=new String[nf];
            in=new int[n];
            a[0]=st;
            while(i<nf){
                for(m=0;m<n;m++){
                    in[m]=n;
                }j=0;
                while(j<n){
                    k=(int)(n*Math.random());

                    for(m=0;m<=j;m++){
                        if(k==in[m]){
                            flag=false;
                            break;          
                        }
                    }
                    if(flag==true){
                        in[j++]=k;
                    }flag=true;
                }s="";
                for(j=0;j<n;j++){
                    s+=l[in[j]];
                }

                //Removing same words
                for(m=0;m<=i;m++){
                        if(s.equalsIgnoreCase(a[m])){
                            flag=false;
                            break;          
                        }
                    }
                    if(flag==true){
                        a[i++]=s;
                        p.println(i+" . "+a[i-1]);
                    }flag=true;

            }

    }
}

我们可以用阶乘来计算有多少字符串以某个字母开头。

示例:取输入abcd。(3!) == 6个字符串将以abcd中的每个字母开头。

static public int facts(int x){
    int sum = 1;
    for (int i = 1; i < x; i++) {
        sum *= (i+1);
    }
    return sum;
}

public static void permutation(String str) {
    char[] str2 = str.toCharArray();
    int n = str2.length;
    int permutation = 0;
    if (n == 1) {
        System.out.println(str2[0]);
    } else if (n == 2) {
        System.out.println(str2[0] + "" + str2[1]);
        System.out.println(str2[1] + "" + str2[0]);
    } else {
        for (int i = 0; i < n; i++) {
            if (true) {
                char[] str3 = str.toCharArray();
                char temp = str3[i];
                str3[i] = str3[0];
                str3[0] = temp;
                str2 = str3;
            }

            for (int j = 1, count = 0; count < facts(n-1); j++, count++) {
                if (j != n-1) {
                    char temp1 = str2[j+1];
                    str2[j+1] = str2[j];
                    str2[j] = temp1;
                } else {
                    char temp1 = str2[n-1];
                    str2[n-1] = str2[1];
                    str2[1] = temp1;
                    j = 1;
                } // end of else block
                permutation++;
                System.out.print("permutation " + permutation + " is   -> ");
                for (int k = 0; k < n; k++) {
                    System.out.print(str2[k]);
                } // end of loop k
                System.out.println();
            } // end of loop j
        } // end of loop i
    }
}

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);
            }
        }

Java中一个非常基本的解决方案是使用递归+设置(以避免重复)，如果你想存储和返回解决方案字符串:

public static Set<String> generatePerm(String input)
{
    Set<String> set = new HashSet<String>();
    if (input == "")
        return set;

    Character a = input.charAt(0);

    if (input.length() > 1)
    {
        input = input.substring(1);

        Set<String> permSet = generatePerm(input);

        for (String x : permSet)
        {
            for (int i = 0; i <= x.length(); i++)
            {
                set.add(x.substring(0, i) + a + x.substring(i));
            }
        }
    }
    else
    {
        set.add(a + "");
    }
    return set;
}