其中一个简单的解决方案是使用两个指针继续递归地交换字符。

public static void main(String[] args)
{
    String str="abcdefgh";
    perm(str);
}
public static void perm(String str)
{  char[] char_arr=str.toCharArray();
    helper(char_arr,0);
}
public static void helper(char[] char_arr, int i)
{
    if(i==char_arr.length-1)
    {
        // print the shuffled string 
            String str="";
            for(int j=0; j<char_arr.length; j++)
            {
                str=str+char_arr[j];
            }
            System.out.println(str);
    }
    else
    {
    for(int j=i; j<char_arr.length; j++)
    {
        char tmp = char_arr[i];
        char_arr[i] = char_arr[j];
        char_arr[j] = tmp;
        helper(char_arr,i+1);
        char tmp1 = char_arr[i];
        char_arr[i] = char_arr[j];
        char_arr[j] = tmp1;
    }
}
}

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
    }
}

其中一个简单的解决方案是使用两个指针继续递归地交换字符。

public static void main(String[] args)
{
    String str="abcdefgh";
    perm(str);
}
public static void perm(String str)
{  char[] char_arr=str.toCharArray();
    helper(char_arr,0);
}
public static void helper(char[] char_arr, int i)
{
    if(i==char_arr.length-1)
    {
        // print the shuffled string 
            String str="";
            for(int j=0; j<char_arr.length; j++)
            {
                str=str+char_arr[j];
            }
            System.out.println(str);
    }
    else
    {
    for(int j=i; j<char_arr.length; j++)
    {
        char tmp = char_arr[i];
        char_arr[i] = char_arr[j];
        char_arr[j] = tmp;
        helper(char_arr,i+1);
        char tmp1 = char_arr[i];
        char_arr[i] = char_arr[j];
        char_arr[j] = tmp1;
    }
}
}

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());
        }
    }
}

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;
}

//循环'整个字符数组，并保持'i'作为你的排列的基础，并像你交换[ab, ba]一样继续寻找组合

public class Permutation {
    //Act as a queue
    private List<Character> list;
    //To remove the duplicates
    private Set<String> set = new HashSet<String>();

    public Permutation(String s) {
        list = new LinkedList<Character>();
        int len = s.length();
        for(int i = 0; i < len; i++) {
            list.add(s.charAt(i));
        }
    }

    public List<String> getStack(Character c, List<Character> list) {
        LinkedList<String> stack = new LinkedList<String>();
        stack.add(""+c);
        for(Character ch: list) {
            stack.add(""+ch);
        }

        return stack;
    }

    public String printCombination(String s1, String s2) {
        //S1 will be a single character
        StringBuilder sb = new StringBuilder();
        String[] strArr = s2.split(",");
        for(String s: strArr) {
            sb.append(s).append(s1);
            sb.append(",");
        }       
        for(String s: strArr) {
            sb.append(s1).append(s);
            sb.append(",");
        }

        return sb.toString();
    }

    public void printPerumtation() {
        int cnt = list.size();

        for(int i = 0; i < cnt; i++) {
            Character c = list.get(0);
            list.remove(0);
            List<String> stack = getStack(c, list);

            while(stack.size() > 1) {
                //Remove the top two elements
                String s2 = stack.remove(stack.size() - 1);
                String s1 = stack.remove(stack.size() - 1);
                String comS = printCombination(s1, s2);
                stack.add(comS);
            }

            String[] perms = (stack.remove(0)).split(",");
            for(String perm: perms) {
                set.add(perm);
            }

            list.add(c);
        }

        for(String s: set) {
            System.out.println(s);
        }
    }
}