python实现

def getPermutation(s, prefix=''):
        if len(s) == 0:
                print prefix
        for i in range(len(s)):
                getPermutation(s[0:i]+s[i+1:len(s)],prefix+s[i] )



getPermutation('abcd','')

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);
            }
        }

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));
    }   
}

没有递归的Java实现

public Set<String> permutate(String s){
    Queue<String> permutations = new LinkedList<String>();
    Set<String> v = new HashSet<String>();
    permutations.add(s);

    while(permutations.size()!=0){
        String str = permutations.poll();
        if(!v.contains(str)){
            v.add(str);
            for(int i = 0;i<str.length();i++){
                String c = String.valueOf(str.charAt(i));
                permutations.add(str.substring(i+1) + c +  str.substring(0,i));
            }
        }
    }
    return v;
}

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]。

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
    }
}