String 生成字符串的所有可能排列的列表
如何生成长度介于x和y字符之间的字符串的所有可能排列的列表,其中包含一个可变的字符列表String 生成字符串的所有可能排列的列表,string,language-agnostic,cross-platform,String,Language Agnostic,Cross Platform,如何生成长度介于x和y字符之间的字符串的所有可能排列的列表,其中包含一个可变的字符列表 任何语言都可以,但它应该是可移植的。有几种方法可以做到这一点。常用方法使用递归、记忆或动态编程。基本思想是生成一个长度为1的所有字符串的列表,然后在每次迭代中,对于上一次迭代中生成的所有字符串,分别添加与字符串中每个字符连接的字符串。(下面代码中的变量索引跟踪上一次迭代和下一次迭代的开始) 一些伪代码: list = originalString.split('') index = (0,0) list =
任何语言都可以,但它应该是可移植的。有几种方法可以做到这一点。常用方法使用递归、记忆或动态编程。基本思想是生成一个长度为1的所有字符串的列表,然后在每次迭代中,对于上一次迭代中生成的所有字符串,分别添加与字符串中每个字符连接的字符串。(下面代码中的变量索引跟踪上一次迭代和下一次迭代的开始) 一些伪代码:
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:
list.add(s + c)
然后,您需要删除所有长度小于x的字符串,它们将是列表中的第一个(x-1)*len(originalString)项。我刚刚在Ruby中快速完成了这一操作:
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
def perms(x,y,可能的_字符)
全部=[“”]
当前_数组=all.clone
1.高达(y){迭代|
下一个数组=[]
当前_数组。每个{124;字符串|
可能的_字符。每个{124; c|
value=string+c
下一个数组。插入下一个数组。长度,值
all.insert all.length、value
}
}
当前数组=下一个数组
}
如果{| string | string.length
您可能会研究内置置换类型函数的语言API,您可能会编写更优化的代码,但如果数量如此之多,我不确定有多少方法可以获得大量结果
无论如何,代码背后的想法是从长度为0的字符串开始,然后跟踪所有长度为Z的字符串,其中Z是迭代中的当前大小。然后,遍历每个字符串并将每个字符附加到每个字符串上。最后,删除所有低于x阈值的项并返回结果
我没有使用可能毫无意义的输入(空字符列表、奇怪的x和y值等)来测试它。我不知道您为什么首先要这样做。任何中等大小的x和y值的结果集都将是巨大的,并且会随着x和/或y变大而呈指数增长 假设您的可能字符集是字母表中的26个小写字母,您要求应用程序生成长度为5的所有排列。假设你没有用完内存,你会得到11881376(即26的5次方)字符串。将该长度增加到6,您将获得308915776个字符串。这些数字很快就变得非常庞大 下面是我用Java编写的一个解决方案。您需要提供两个运行时参数(对应于x和y)。玩得开心
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);
}
}
}
}
公共类生成项{
公共静态void main(字符串[]args){
int lower=Integer.parseInt(args[0]);
int upper=Integer.parseInt(args[1]);
如果(上<下| |上==0 | |下==0){
系统出口(0);
}
对于(int-length=lower;length您将获得大量字符串,这是肯定的
其中x和y是定义它们的方式,r是我们从中选择的字符数——如果我理解正确的话。你应该根据需要生成它们,而不是草率地说,生成一个powerset,然后过滤字符串的长度
下面的内容肯定不是生成这些内容的最佳方式,但它仍然是一个有趣的旁白
Knuth(第4卷,第2分册,7.2.1.3)告诉我们,(s,t)-组合相当于一次取t的s+1事物的重复-(s,t)-组合是Knuth使用的表示法,它等于。我们可以通过首先以二进制形式生成每个(s,t)-组合(长度为(s+t))并计算每个1左边0的数量
1000011101-->成为ruby中的置换:{0,3,4,4,4,1}:
str = "a"
100_000_000.times {puts str.next!}
它相当快,但需要一些时间(当然,如果短字符串对您不感兴趣,您可以从“aaaaaaaa”开始
不过,我可能误解了实际的问题——在其中一篇文章中,它听起来好像你只需要一个bruteforce字符串库,但在主要问题中,它听起来像你需要排列一个特定的字符串
您的问题与此类似:(列出所有整数,其中没有一个数字重复,这导致许多语言解决了这个问题,ocaml使用排列,而java使用另一种解决方案)。这是Mike的Ruby版本到Common Lisp的翻译:
(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)))))
虽然这并不能准确回答您的问题,但有一种方法可以从许多相同长度的字符串中生成字母的每个排列:例如,如果您的单词是“coffee”、“joomla”和“moodle”,您可以期望输出“coodle”、“joodee”、“joffle”等
基本上,组合数是(单词数)到(每个单词的字母数)的幂。因此,在0和组合数-1之间选择一个随机数,将该数字转换为基数(单词数),然后使用该数字的每个数字作为从中提取下一个字母的指标
在上面的例子中,3个单词,6个字母=729个组合。选择一个随机数:465。转换为基数3:122020。从单词1中取第一个字母,从单词2中取第二个字母,从单词2中取第三个字母,从单词0中取第四个字母……然后你得到了……“joofle”
如果你想要所有的排列,只需从0循环到728。当然,如果你只是选择一个随机值,一个简单得多、不那么混乱的方法就是循环字母。如果你想要所有的排列,这个方法可以避免递归,而且它让你看起来像懂数学(tm)
如果组合的数量过多,您可以将其分解为一系列较小的单词,并在末尾将它们连接起来。中的递归解决方案
(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)))))
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 ArrayList CalculateWordPermutations(string[] letters, ArrayList words, int index)
{
bool finished = true;
ArrayList newWords = new ArrayList();
if (words.Count == 0)
{
foreach (string letter in letters)
{
words.Add(letter);
}
}
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];
newWords.Add(newWord);
}
}
}
foreach (string newWord in newWords)
{
words.Add(newWord);
}
if(finished == false)
{
CalculateWordPermutations(letters, words, words.Count - newWords.Count);
}
return words;
}
string[] letters = new string[]{"a","b","c"};
ArrayList words = CalculateWordPermutations(letters, new ArrayList(), 0);
my @result = ("a" .. "zzzz");
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);
}
}
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;
}
}
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) {
ret_set.add(c+s);
}
return ret_set;
}
public static HashSet<String> all_perm(String a) {
HashSet<String> set = new HashSet<String>();
if(a.length() == 1) {
set.add(a);
} else {
for(int i=0; i<a.length(); i++) {
set.addAll(concat(a.charAt(i)+"", all_perm(a.substring(0, i)+a.substring(i+1, a.length()))));
}
}
return set;
}
}
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)
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;
}
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>();
ret.add(str);
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;
}
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()
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;
}
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")
public List<string> Permutations(char[] chars)
{
List<string> words = new List<string>();
words.Add(chars[0].ToString());
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
words.Add(nstr);
}
}
}
return words;
}
//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()));
}
}
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
#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 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
from itertools import permutations
s = 'ABCDEF'
p = [''.join(x) for x in permutations(s)]
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)
}