生成字符串的所有可能排列的列表

37 个答案:

list = originalString.split('')
index = (0,0)
list = [""]
for iteration n in 1 to y:
index = (index[1], len(list))
for string s in list.subset(index[0] to end):
for character c in originalString:


#include <stdio.h>
#include <string.h>

void swap(char *a, char *b) {
char temp;
temp = *a;
*a = *b;
*b = temp;
}

void print(char *a, int i, int n) {
int j;
if(i == n) {
printf("%s\n", a);
} else {
for(j = i; j <= n; j++) {
swap(a + i, a + j);
print(a, i + 1, n);
swap(a + i, a + j);
}
}
}

int main(void) {
char a[100];
gets(a);
print(a, 0, strlen(a) - 1);
return 0;
}


\sum_{i=x}^y{\frac{r!}{{(r-i)}!}} http://www.codecogs.com/eq.latex?%5Csum_%7Bi=x%7D%5Ey%20%7B%20%5Cfrac%7Br!%7D%7B%7B(r-i)%7D!%7D%20%7D

Knuth（第4卷，第2节，7.2.1.3）告诉我们，（s，t） - 组合相当于重复时采用的s + 1个事物 - （s，t） - 组合是Knuth使用的符号等于{t \choose {s+t} http://www.codecogs.com/eq.latex?%7Bt%20%5Cchoose%20%7Bs+t%7D%7D。我们可以通过首先以二进制形式生成每个（s，t）组合（因此，长度（s + t））并计算每个的左边的0的数量来计算出来。

10001000011101 - ＆gt;成为排列：{0,3,4,4,4,1}

def nextPermutation(perm):
k0 = None
for i in range(len(perm)-1):
if perm[i]<perm[i+1]:
k0=i
if k0 == None:
return None

l0 = k0+1
for i in range(k0+1, len(perm)):
if perm[k0] < perm[i]:
l0 = i

perm[k0], perm[l0] = perm[l0], perm[k0]
perm[k0+1:] = reversed(perm[k0+1:])
return perm

perm=list("12345")
while perm:
print perm
perm = nextPermutation(perm)


    static public IEnumerable<string> permute(string word)
{
if (word.Length > 1)
{

char character = word[0];
foreach (string subPermute in permute(word.Substring(1)))
{

for (int index = 0; index <= subPermute.Length; index++)
{
string pre = subPermute.Substring(0, index);
string post = subPermute.Substring(index);

if (post.Contains(character))
continue;

yield return pre + character + post;
}

}
}
else
{
yield return word;
}
}


public class permute {

static void permute(int level, String permuted,
boolean used[], String original) {
int length = original.length();
if (level == length) {
System.out.println(permuted);
} else {
for (int i = 0; i < length; i++) {
if (!used[i]) {
used[i] = true;
permute(level + 1, permuted + original.charAt(i),
used, original);
used[i] = false;
}
}
}
}

public static void main(String[] args) {
String s = "hello";
boolean used[] = {false, false, false, false, false};
permute(0, "", used, s);
}
}


#include <string>
#include <iostream>

template<typename Consume>
void permutations(std::string s, Consume consume, std::size_t start = 0) {
if (start == s.length()) consume(s);
for (std::size_t i = start; i < s.length(); i++) {
std::swap(s[start], s[i]);
permutations(s, consume, start + 1);
}
}

int main(void) {
std::string s = "abcd";
permutations(s, [](std::string s) {
std::cout << s << std::endl;
});
}


def perms(x, y, possible_characters)
all = [""]
current_array = all.clone
1.upto(y) { |iteration|
next_array = []
current_array.each { |string|
possible_characters.each { |c|
value = string + c
next_array.insert next_array.length, value
all.insert all.length, value
}
}
current_array = next_array
}
all.delete_if { |string| string.length < x }
end


C ++中的递归解决方案

int main (int argc, char * const argv[]) {
string s = "sarp";
bool used [4];
permute(0, "", used, s);
}

void permute(int level, string permuted, bool used [], string &original) {
int length = original.length();

if(level == length) { // permutation complete, display
cout << permuted << endl;
} else {
for(int i=0; i<length; i++) { // try to add an unused character
if(!used[i]) {
used[i] = true;
permute(level+1, original[i] + permuted, used, original); // find the permutations starting with this string
used[i] = false;
}
}
}


public static void main(String[] args) {

for (String str : permStr("ABBB")){
System.out.println(str);
}
}

static Vector<String> permStr(String str){

if (str.length() == 1){
Vector<String> ret = new Vector<String>();
return ret;
}

char start = str.charAt(0);
Vector<String> endStrs = permStr(str.substring(1));
Vector<String> newEndStrs = new Vector<String>();
for (String endStr : endStrs){
for (int j = 0; j <= endStr.length(); j++){
if (endStr.substring(0, j).endsWith(String.valueOf(start)))
break;
newEndStrs.add(endStr.substring(0, j) + String.valueOf(start) + endStr.substring(j));
}
}
return newEndStrs;
}


public ArrayList CalculateWordPermutations(string[] letters, ArrayList words, int index)
{
bool finished = true;
ArrayList newWords = new ArrayList();
if (words.Count == 0)
{
foreach (string letter in letters)
{
}
}

for(int j=index; j<words.Count; j++)
{
string word = (string)words[j];
for(int i =0; i<letters.Length; i++)
{
if(!word.Contains(letters[i]))
{
finished = false;
string newWord = (string)word.Clone();
newWord += letters[i];
}
}
}

foreach (string newWord in newWords)
{
}

if(finished  == false)
{
CalculateWordPermutations(letters, words, words.Count - newWords.Count);
}
return words;
}


<强>调用

string[] letters = new string[]{"a","b","c"};
ArrayList words = CalculateWordPermutations(letters, new ArrayList(), 0);


...这里是C版：

void permute(const char *s, char *out, int *used, int len, int lev)
{
if (len == lev) {
out[lev] = '\0';
puts(out);
return;
}

int i;
for (i = 0; i < len; ++i) {
if (! used[i])
continue;

used[i] = 1;
out[lev] = s[i];
permute(s, out, used, len, lev + 1);
used[i] = 0;
}
return;
}


(defun perms (x y original-string)
(loop with all = (list "")
with current-array = (list "")
for iteration from 1 to y
do (loop with next-array = nil
for string in current-array
do (loop for c across original-string
for value = (concatenate 'string string (string c))
do (push value next-array)
(push value all))
(setf current-array (reverse next-array)))
finally (return (nreverse (delete-if #'(lambda (el) (< (length el) x)) all)))))


(defun perms (x y original-string)
(loop repeat y
collect (loop for string in (or (car (last sets)) (list ""))
append (loop for c across original-string
collect (concatenate 'string string (string c)))) into sets
finally (return (loop for set in sets
append (loop for el in set when (>= (length el) x) collect el)))))


Ruby回答有效：

class String
def each_char_with_index
0.upto(size - 1) do |index|
yield(self[index..index], index)
end
end
def remove_char_at(index)
return self[1..-1] if index == 0
self[0..(index-1)] + self[(index+1)..-1]
end
end

def permute(str, prefix = '')
if str.size == 0
puts prefix
return
end
str.each_char_with_index do |char, index|
permute(str.remove_char_at(index), prefix + char)
end
end

# example
# permute("abc")


my @result = ("a" .. "zzzz");


//call it as permut("",str);

public void permut(String str1,String str2){
if(str2.length() != 0){
char ch = str2.charAt(0);
for(int i = 0; i <= str1.length();i++)
permut(str1.substring(0,i) + ch + str1.substring(i,str1.length()),
str2.substring(1,str2.length()));
}else{
System.out.println(str1);
}
}


//call it as permut("",str);

public void permut(String str1,String str2){
if(str2.length() > 1){
char ch = str2.charAt(0);
for(int i = 0; i <= str1.length();i++)
permut(str1.substring(0,i) + ch + str1.substring(i,str1.length()),
str2.substring(1,str2.length()));
}else{
char ch = str2.charAt(0);
for(int i = 0; i <= str1.length();i++)
System.out.println(str1.substring(0,i) + ch +    str1.substring(i,str1.length()),
str2.substring(1,str2.length()));
}
}


1. 给定单个源数组，生成第一组新数组，这些数组依次交换第一个元素和每个后续元素，每次都不会干扰其他元素。 例如：给出1234，生成1234,2134,3214,4231。

2. 使用上一遍中的每个数组作为新传递的种子， 但不是交换第一个元素，而是将第二个元素与每个后续元素交换。此外，这次不要在输出中包含原始数组。

3. 重复步骤2直至完成。

4. 以下是代码示例：

function oxe_perm(src, depth, index)
{
var perm = src.slice();     // duplicates src.
perm = perm.split("");
perm[depth] = src[index];
perm[index] = src[depth];
perm = perm.join("");
return perm;
}

function oxe_permutations(src)
{
out = new Array();

out.push(src);

for (depth = 0; depth < src.length; depth++) {
var numInPreviousPass = out.length;
for (var m = 0; m < numInPreviousPass; ++m) {
for (var n = depth + 1; n < src.length; ++n) {
out.push(oxe_perm(out[m], depth, n));
}
}
}

return out;
}


import java.util.*;

public class all_subsets {
public static void main(String[] args) {
String a = "abcd";
for(String s: all_perm(a)) {
System.out.println(s);
}
}

public static Set<String> concat(String c, Set<String> lst) {
HashSet<String> ret_set = new HashSet<String>();
for(String s: lst) {
}
return ret_set;
}

public static HashSet<String> all_perm(String a) {
HashSet<String> set = new HashSet<String>();
if(a.length() == 1) {
} else {
for(int i=0; i<a.length(); i++) {
}
}
return set;
}
}


public class GeneratePermutations {
public static void main(String[] args) {
int lower = Integer.parseInt(args[0]);
int upper = Integer.parseInt(args[1]);

if (upper < lower || upper == 0 || lower == 0) {
System.exit(0);
}

for (int length = lower; length <= upper; length++) {
generate(length, "");
}
}

private static void generate(int length, String partial) {
if (length <= 0) {
System.out.println(partial);
} else {
for (char c = 'a'; c <= 'z'; c++) {
generate(length - 1, partial + c);
}
}
}
}


* VB会为你做到这一点。

**您可以稍微更改NextPerm以使其不必要，但它更清晰。

Option Explicit

Function NextPerm(Perm() As Long, Stack() As Long, Level As Long) As Boolean
Dim N As Long
If Level = 2 Then
Swap Perm(1), Perm(2)
Level = 3
Else
While Stack(Level) = Level - 1
Stack(Level) = 0
If Level = UBound(Stack) Then Exit Function
Level = Level + 1
Wend
Stack(Level) = Stack(Level) + 1
If Level And 1 Then N = 1 Else N = Stack(Level)
Swap Perm(N), Perm(Level)
Level = 2
End If
NextPerm = True
End Function

Sub Swap(A As Long, B As Long)
A = A Xor B
B = A Xor B
A = A Xor B
End Sub

'This is just for testing.
Private Sub Form_Paint()
Const Max = 8
Dim A(1 To Max) As Long, I As Long
Dim S(3 To Max) As Long, J As Long
Dim Test As New Collection, T As String
For I = 1 To UBound(A)
A(I) = I
Next
Cls
ScaleLeft = 0
J = 2
Do
If CurrentY + TextHeight("0") > ScaleHeight Then
ScaleLeft = ScaleLeft - TextWidth(" 0 ") * (UBound(A) + 1)
CurrentY = 0
CurrentX = 0
End If
T = vbNullString
For I = 1 To UBound(A)
Print A(I);
T = T & Hex(A(I))
Next
Print
Loop While NextPerm(A, S, J)
J = 1
For I = 2 To UBound(A)
J = J * I
Next
If J <> Test.Count Then Stop
End Sub


str = "a"
100_000_000.times {puts str.next!}


['0000', '0001', '0010', '0011', '0100', '0101', '0110', '0111', '1000', '1001', '1010', '1011', '1100', '1101', '1110', '1111']

def generate_permutations(chars = 4) :

#modify if in need!
allowed_chars = [
'0',
'1',
]

status = []
for tmp in range(chars) :
status.append(0)

last_char = len(allowed_chars)

rows = []
for x in xrange(last_char ** chars) :
rows.append("")
for y in range(chars - 1 , -1, -1) :
key = status[y]
rows[x] = allowed_chars[key] + rows[x]

for pos in range(chars - 1, -1, -1) :
if(status[pos] == last_char - 1) :
status[pos] = 0
else :
status[pos] += 1
break;

return rows

import sys

print generate_permutations()


class Permutation {

/* runtime -O(n) for generating nextPermutaion
* and O(n*n!) for generating all n! permutations with increasing sorted array as start
* return true, if there exists next lexicographical sequence
* e.g [a,b,c],3-> true, modifies array to [a,c,b]
* e.g [c,b,a],3-> false, as it is largest lexicographic possible */
public static boolean nextPermutation(char[] seq, int len) {
// 1
if (len <= 1)
return false;// no more perm
// 2: Find last j such that seq[j] <= seq[j+1]. Terminate if no such j exists
int j = len - 2;
while (j >= 0 && seq[j] >= seq[j + 1]) {
--j;
}
if (j == -1)
return false;// no more perm
// 3: Find last l such that seq[j] <= seq[l], then exchange elements j and l
int l = len - 1;
while (seq[j] >= seq[l]) {
--l;
}
swap(seq, j, l);
// 4: Reverse elements j+1 ... count-1:
reverseSubArray(seq, j + 1, len - 1);
// return seq, add store next perm

return true;
}
private static void swap(char[] a, int i, int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}

private static void reverseSubArray(char[] a, int lo, int hi) {
while (lo < hi) {
swap(a, lo, hi);
++lo;
--hi;
}
}
public static void main(String[] args) {
String str = "abcdefg";
char[] array = str.toCharArray();
Arrays.sort(array);
int cnt=0;
do {
System.out.println(new String(array));
cnt++;
}while(nextPermutation(array, array.length));
System.out.println(cnt);//5040=7!
}
//if we use "bab"-> "abb", "bab", "bba", 3(#permutations)
}


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;
}


python中的递归解决方案。这段代码的好处是它导出一个字典，键作为字符串，所有可能的排列作为值。包含所有可能的字符串长度，因此实际上，您正在创建超集。

• 使用较小的字符串调用自身
• 返回特定字符串的排列列表（如果已有）。如果返回自身，这些将用于附加到角色并创建更新的排列。

perms = {}
def perm(input_string):
global perms
if input_string in perms:
return perms[input_string] # This will send a list of all permutations
elif len(input_string) == 2:
perms[input_string] = [input_string, input_string[-1] + input_string [-2]]
return perms[input_string]
else:
perms[input_string] = []
for index in range(0, len(input_string)):
new_string = input_string[0:index] + input_string[index +1:]
perm(new_string)
for entries in perms[new_string]:
perms[input_string].append(input_string[index] + entries)
return perms[input_string]


def gen( x,y,list): #to generate all strings inserting y at different positions
list = []
list.append( y+x )
for i in range( len(x) ):
list.append( func(x,0,i) + y + func(x,i+1,len(x)-1) )
return list

def func( x,i,j ): #returns x[i..j]
z = ''
for i in range(i,j+1):
z = z+x[i]
return z

def perm( x , length , list ): #perm function
if length == 1 : # base case
list.append( x[len(x)-1] )
return list
else:
lists = perm( x , length-1 ,list )
lists_temp = lists #temporarily storing the list
lists = []
for i in range( len(lists_temp) ) :
list_temp = gen(lists_temp[i],x[length-2],lists)
lists += list_temp
return lists


package namo.algorithms;

import java.util.Scanner;

public class Permuations {

public static int totalPermutationsCount = 0;
public static void main(String[] args) {

Scanner sc = new Scanner(System.in);
System.out.println("input string : ");
String inputString = sc.nextLine();
System.out.println("given input String ==> "+inputString+ " :: length is = "+inputString.length());
findPermuationsOfString(-1, inputString);
System.out.println("**************************************");
System.out.println("total permutation strings ==> "+totalPermutationsCount);
}

public  static void findPermuationsOfString(int fixedIndex, String inputString) {
int currentIndex = fixedIndex +1;

for (int i = currentIndex; i < inputString.length(); i++) {
//swap elements and call the findPermuationsOfString()

char[] carr = inputString.toCharArray();
char tmp = carr[currentIndex];
carr[currentIndex] = carr[i];
carr[i] = tmp;
inputString =  new String(carr);

//System.out.println("chat At : current String ==> "+inputString.charAt(currentIndex));
if(currentIndex == inputString.length()-1) {
totalPermutationsCount++;
System.out.println("permuation string ==> "+inputString);
} else {
//System.out.println("in else block>>>>");
findPermuationsOfString(currentIndex, inputString);
char[] rarr = inputString.toCharArray();
char rtmp = carr[i];
carr[i] = carr[currentIndex];
carr[currentIndex] = rtmp;
inputString =  new String(carr);
}
}
}


}

def permutation(str)
posibilities = []
str.split('').each do |char|
if posibilities.size == 0
posibilities[0] = char.downcase
posibilities[1] = char.upcase
else
posibilities_count = posibilities.length
posibilities = posibilities + posibilities
posibilities_count.times do |i|
posibilities[i] += char.downcase
posibilities[i+posibilities_count] += char.upcase
end
end
end
posibilities
end


UncommonsMaths中有一个迭代的Java实现（适用于对象列表）：

/**
* Generate the indices into the elements array for the next permutation. The
* algorithm is from Kenneth H. Rosen, Discrete Mathematics and its
* Applications, 2nd edition (NY: McGraw-Hill, 1991), p. 284)
*/
private void generateNextPermutationIndices()
{
if (remainingPermutations == 0)
{
throw new IllegalStateException("There are no permutations " +
"remaining. Generator must be reset to continue using.");
}
else if (remainingPermutations < totalPermutations)
{
// Find largest index j with
// permutationIndices[j] < permutationIndices[j + 1]
int j = permutationIndices.length - 2;
while (permutationIndices[j] > permutationIndices[j + 1])
{
j--;
}

// Find index k such that permutationIndices[k] is smallest integer
// greater than permutationIndices[j] to the right
// of permutationIndices[j].
int k = permutationIndices.length - 1;
while (permutationIndices[j] > permutationIndices[k])
{
k--;
}

// Interchange permutation indices.
int temp = permutationIndices[k];
permutationIndices[k] = permutationIndices[j];
permutationIndices[j] = temp;

// Put tail end of permutation after jth position in increasing order.
int r = permutationIndices.length - 1;
int s = j + 1;

while (r > s)
{
temp = permutationIndices[s];
permutationIndices[s] = permutationIndices[r];
permutationIndices[r] = temp;
r--;
s++;
}
}
--remainingPermutations;
}

/**
* Generate the next permutation and return a list containing
* the elements in the appropriate order.  This overloaded method
* allows the caller to provide a list that will be used and returned.
* The purpose of this is to improve performance when iterating over
* permutations.  If the {@link #nextPermutationAsList()} method is
* used it will create a new list every time.  When iterating over
* permutations this will result in lots of short-lived objects that
* have to be garbage collected.  This method allows a single list
* instance to be reused in such circumstances.
* @param destination Provides a list to use to create the
* permutation.  This is the list that will be returned, once
* it has been filled with the elements in the appropriate order.
* @return The next permutation as a list.
*/
public List<T> nextPermutationAsList(List<T> destination)
{
generateNextPermutationIndices();
// Generate actual permutation.
destination.clear();
for (int i : permutationIndices)
{
}
return destination;
}


public static String insertCharAt(String s, int index, char c) {
StringBuffer sb = new StringBuffer(s);
StringBuffer sbb = sb.insert(index, c);
return sbb.toString();
}

public static ArrayList<String> getPerm(String s, int index) {
ArrayList<String> perm = new ArrayList<String>();

if (index == s.length()-1) {
return perm;
}

ArrayList<String> p = getPerm(s, index+1);
char c = s.charAt(index);

for(String pp : p) {
for (int idx=0; idx<pp.length()+1; idx++) {
String ss = insertCharAt(pp, idx, c);
}
}

return perm;
}

public static void testGetPerm(String s) {
ArrayList<String> perm = getPerm(s,0);
System.out.println(s+" --> total permutation are :: "+perm.size());
System.out.println(perm.toString());
}


func StringPermutations(inputStr string) (permutations []string) {
for i := 0; i < len(inputStr); i++ {
inputStr = inputStr[1:] + inputStr[0:1]
if len(inputStr) <= 2 {
permutations = append(permutations, inputStr)
continue
}
leftPermutations := StringPermutations(inputStr[0 : len(inputStr)-1])
for _, leftPermutation := range leftPermutations {
permutations = append(permutations, leftPermutation+inputStr[len(inputStr)-1:])
}
}
return
}


pythonic解决方案：

from itertools import permutations
s = 'ABCDEF'
p = [''.join(x) for x in permutations(s)]


public class AllPermutationsOfString {
public static void stringPermutations(String newstring, String remaining) {
if(remaining.length()==0)
System.out.println(newstring);

for(int i=0; i<remaining.length(); i++) {
String newRemaining = remaining.replaceFirst(remaining.charAt(i)+"", "");
stringPermutations(newstring+remaining.charAt(i), newRemaining);
}
}

public static void main(String[] args) {
String string = "abc";
AllPermutationsOfString.stringPermutations("", string);
}


}

c＃iterative：

public List<string> Permutations(char[] chars)
{
List<string> words = new List<string>();
for (int i = 1; i < chars.Length; ++i)
{
int currLen = words.Count;
for (int j = 0; j < currLen; ++j)
{
var w = words[j];
for (int k = 0; k <= w.Length; ++k)
{
var nstr = w.Insert(k, chars[i].ToString());
if (k == 0)
words[j] = nstr;
else