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
    }
}

这里有一个优雅的，非递归的O(n!)解:

public static StringBuilder[] permutations(String s) {
        if (s.length() == 0)
            return null;
        int length = fact(s.length());
        StringBuilder[] sb = new StringBuilder[length];
        for (int i = 0; i < length; i++) {
            sb[i] = new StringBuilder();
        }
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            int times = length / (i + 1);
            for (int j = 0; j < times; j++) {
                for (int k = 0; k < length / times; k++) {
                    sb[j * length / times + k].insert(k, ch);
                }
            }
        }
        return sb;
    }

下面是一个简单的Java递归解决方案:

public static ArrayList<String> permutations(String s) {
    ArrayList<String> out = new ArrayList<String>();
    if (s.length() == 1) {
        out.add(s);
        return out;
    }
    char first = s.charAt(0);
    String rest = s.substring(1);
    for (String permutation : permutations(rest)) {
        out.addAll(insertAtAllPositions(first, permutation));
    }
    return out;
}
public static ArrayList<String> insertAtAllPositions(char ch, String s) {
    ArrayList<String> out = new ArrayList<String>();
    for (int i = 0; i <= s.length(); ++i) {
        String inserted = s.substring(0, i) + ch + s.substring(i);
        out.add(inserted);
    }
    return out;
}

下面是一个java实现:

/* All Permutations of a String */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Complexity O(n*n!) */
class Ideone
{
     public static ArrayList<String> strPerm(String str, ArrayList<String> list)
     {
        int len = str.length();
        if(len==1){
            list.add(str);
            return list;
        }

        list = strPerm(str.substring(0,len-1),list);
        int ls = list.size();
        char ap = str.charAt(len-1);
        for(int i=0;i<ls;i++){
            String temp = list.get(i);
            int tl = temp.length();
            for(int j=0;j<=tl;j++){
                list.add(temp.substring(0,j)+ap+temp.substring(j,tl));  
            }
        }

        while(true){
            String temp = list.get(0);
            if(temp.length()<len)
                list.remove(temp);
            else
                break;
        }

        return list;
    }

    public static void main (String[] args) throws java.lang.Exception
    {
        String str = "abc";
        ArrayList<String> list = new ArrayList<>();

        list = strPerm(str,list);
        System.out.println("Total Permutations : "+list.size());
        for(int i=0;i<list.size();i++)
            System.out.println(list.get(i));

    }
}

http://ideone.com/nWPb3k

使用递归。

当输入是空字符串时，唯一的排列就是空字符串。尝试将字符串中的每个字母作为第一个字母，然后使用递归调用找到其余字母的所有排列。

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

class Permutation {
    private static List<String> permutation(String prefix, String str) {
        List<String> permutations = new ArrayList<>();
        int n = str.length();
        if (n == 0) {
            permutations.add(prefix);
        } else {
            for (int i = 0; i < n; i++) {
                permutations.addAll(permutation(prefix + str.charAt(i), str.substring(i + 1, n) + str.substring(0, i)));
            }
        }
        return permutations;
    }

    public static void main(String[] args) {
        List<String> perms = permutation("", "abcd");

        String[] array = new String[perms.size()];
        for (int i = 0; i < perms.size(); i++) {
            array[i] = perms.get(i);
        }

        int x = array.length;

        for (final String anArray : array) {
            System.out.println(anArray);
        }
    }
}

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)