简单的解决方案，利用swift语言的特点，数组是值类型。

func permutation(chrs: [String], arr: [String], result: inout [[String]]) {
   if arr.count == chrs.count {
       result.append(arr)
       return
   }

   for chr in chrs {
       var arr = arr
       if !arr.contains(chr) {
           arr.append(chr)
           permutation(chrs: chrs, arr: arr, result: &result)
       }
   }
}

func test() {
   var result = [[String]]()
   let chrs = ["a", "b", "c", "d"]
   permutation(chrs: chrs, arr: [], result: &result)
}

复杂度O(n * n!)

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)

//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);
        }
    }
}

这可以通过简单地在前面部分结果的所有位置依次插入字符串的每个字母来迭代完成。

我们以[A]开头，与B连成[BA, AB]，与C连成[CBA, BCA, BAC, CAB等]。

运行时间将是O(n!)，对于测试用例ABCD，它是1 x 2 x 3 x 4。

在上面的乘积中，1是A, 2是B，以此类推。

飞镖示例:

void main() {

  String insertAt(String a, String b, int index)
  {
    return a.substring(0, index) + b + a.substring(index);
  }

  List<String> Permute(String word) {

    var letters = word.split('');

    var p_list = [ letters.first ];

    for (var c in letters.sublist(1)) {

      var new_list = [ ];

      for (var p in p_list)
        for (int i = 0; i <= p.length; i++)
          new_list.add(insertAt(p, c, i));

      p_list = new_list;
    }

    return p_list;
  }

  print(Permute("ABCD"));

}

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
    }
}

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());
        }
    }
}