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)

public class Permutation 
{ 
public static void main(String[] args) 
{ 
    String str = "ABC"; 
    int n = str.length(); 
    Permutation permutation = new Permutation(); 
    permutation.permute(str, 0, n-1); 
} 

/** 
* permutation function 
* @param str string to calculate permutation for 
* @param l starting index 
* @param r end index 
*/
private void permute(String str, int l, int r) 
{ 
    if (l == r) 
        System.out.println(str); 
    else
    { 
        for (int i = l; i <= r; i++) 
        { 
            str = swap(str,l,i); 
            permute(str, l+1, r); 
            str = swap(str,l,i); 
        } 
    } 
} 

/** 
* Swap Characters at position 
* @param a string value 
* @param i position 1 
* @param j position 2 
* @return swapped string 
*/
public String swap(String a, int i, int j) 
{ 
    char temp; 
    char[] charArray = a.toCharArray(); 
    temp = charArray[i] ; 
    charArray[i] = charArray[j]; 
    charArray[j] = temp; 
    return String.valueOf(charArray); 
} 

} 

改进的代码相同

    static String permutationStr[];
    static int indexStr = 0;

    static int factorial (int i) {
        if (i == 1)
            return 1;
        else
            return i * factorial(i-1);
    }

    public static void permutation(String str) {
        char strArr[] = str.toLowerCase().toCharArray();
        java.util.Arrays.sort(strArr);

        int count = 1, dr = 1;
        for (int i = 0; i < strArr.length-1; i++){
            if ( strArr[i] == strArr[i+1]) {
                count++;
            } else {
                dr *= factorial(count);
                count = 1;
            }       
        }
        dr *= factorial(count);

        count = factorial(strArr.length) / dr;

        permutationStr = new String[count];

        permutation("", str);

        for (String oneStr : permutationStr){
            System.out.println(oneStr);
        }
    }

    private static void permutation(String prefix, String str) {
        int n = str.length();
        if (n == 0) {
            for (int i = 0; i < indexStr; i++){
                if(permutationStr[i].equals(prefix))
                    return;
            }        
            permutationStr[indexStr++] = prefix;
        } else {
            for (int i = 0; i < n; i++) {
                permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i + 1, n));
            }
        }
    }

以下是我在《破解编程面试》(P54)一书中提出的解决方案:

/**
 * List permutations of a string.
 * 
 * @param s the input string
 * @return  the list of permutations
 */
public static ArrayList<String> permutation(String s) {
    // The result
    ArrayList<String> res = new ArrayList<String>();
    // If input string's length is 1, return {s}
    if (s.length() == 1) {
        res.add(s);
    } else if (s.length() > 1) {
        int lastIndex = s.length() - 1;
        // Find out the last character
        String last = s.substring(lastIndex);
        // Rest of the string
        String rest = s.substring(0, lastIndex);
        // Perform permutation on the rest string and
        // merge with the last character
        res = merge(permutation(rest), last);
    }
    return res;
}

/**
 * @param list a result of permutation, e.g. {"ab", "ba"}
 * @param c    the last character
 * @return     a merged new list, e.g. {"cab", "acb" ... }
 */
public static ArrayList<String> merge(ArrayList<String> list, String c) {
    ArrayList<String> res = new ArrayList<>();
    // Loop through all the string in the list
    for (String s : list) {
        // For each string, insert the last character to all possible positions
        // and add them to the new list
        for (int i = 0; i <= s.length(); ++i) {
            String ps = new StringBuffer(s).insert(i, c).toString();
            res.add(ps);
        }
    }
    return res;
}

字符串"abcd"的运行输出:

第一步:合并[a]和b: [ba, ab] 步骤2:Merge [ba, ab]和c: [cba, bca, bac, cab, acb, abc] 第三步:Merge [cba, bca, bac, cab, acb, abc]和d: [dcba, cdba, cbad, cbca, bdcad

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
    }
}

/** Returns an array list containing all
 * permutations of the characters in s. */
public static ArrayList<String> permute(String s) {
    ArrayList<String> perms = new ArrayList<>();
    int slen = s.length();
    if (slen > 0) {
        // Add the first character from s to the perms array list.
        perms.add(Character.toString(s.charAt(0)));

        // Repeat for all additional characters in s.
        for (int i = 1;  i < slen;  ++i) {

            // Get the next character from s.
            char c = s.charAt(i);

            // For each of the strings currently in perms do the following:
            int size = perms.size();
            for (int j = 0;  j < size;  ++j) {

                // 1. remove the string
                String p = perms.remove(0);
                int plen = p.length();

                // 2. Add plen + 1 new strings to perms.  Each new string
                //    consists of the removed string with the character c
                //    inserted into it at a unique location.
                for (int k = 0;  k <= plen;  ++k) {
                    perms.add(p.substring(0, k) + c + p.substring(k));
                }
            }
        }
    }
    return perms;
}