找出弦的所有排列的优雅方法是什么。例如,ba的排列,将是ba和ab,但更长的字符串,如abcdefgh?是否有Java实现示例?
当前回答
基于Mark Byers的回答,我的python实现:
def permutations(string):
if len(string) == 1:
return [string]
permutations=[]
for i in range(len(string)):
for perm in permutations(string[:i]+string[i+1:]):
permutations.append(string[i] + perm)
return permutations
其他回答
这是一个更快的解决方案,因为它不受字符串连接计算复杂度O(n^2)的影响。另一方面它是无循环的,完全递归的
public static void main(String[] args) {
permutation("ABCDEFGHIJKLMNOPQRSTUVWXYZ");
}
private static void permutation(String str) {
char[] stringArray = str.toCharArray();
printPermutation(stringArray, 0, stringArray.length, 0, 1);
}
private static void printPermutation(char[] string, int loopCounter, int length, int indexFrom, int indexTo) {
// Stop condition
if (loopCounter == length)
return;
/*
When reaching the end of the array:
1- Reset loop indices.
2- Increase length counter.
*/
if (indexTo == length) {
indexFrom = 0;
indexTo = 1;
++loopCounter;
}
// Print.
System.out.println(string);
// Swap from / to indices.
char temp = string[indexFrom];
string[indexFrom] = string[indexTo];
string[indexTo] = temp;
// Go for next iteration.
printPermutation(string, loopCounter, length, ++indexFrom, ++indexTo);
}
简单的递归c++实现如下所示:
#include <iostream>
void generatePermutations(std::string &sequence, int index){
if(index == sequence.size()){
std::cout << sequence << "\n";
} else{
generatePermutations(sequence, index + 1);
for(int i = index + 1 ; i < sequence.size() ; ++i){
std::swap(sequence[index], sequence[i]);
generatePermutations(sequence, index + 1);
std::swap(sequence[index], sequence[i]);
}
}
}
int main(int argc, char const *argv[])
{
std::string str = "abc";
generatePermutations(str, 0);
return 0;
}
输出:
abc
acb
bac
bca
cba
cab
更新
如果想要存储结果,可以将vector作为函数调用的第三个参数传递。此外,如果您只想要唯一的排列,您可以使用集合。
#include <iostream>
#include <vector>
#include <set>
void generatePermutations(std::string &sequence, int index, std::vector <std::string> &v){
if(index == sequence.size()){
//std::cout << sequence << "\n";
v.push_back(sequence);
} else{
generatePermutations(sequence, index + 1, v);
for(int i = index + 1 ; i < sequence.size() ; ++i){
std::swap(sequence[index], sequence[i]);
generatePermutations(sequence, index + 1, v);
std::swap(sequence[index], sequence[i]);
}
}
}
int main(int argc, char const *argv[])
{
std::string str = "112";
std::vector <std::string> permutations;
generatePermutations(str, 0, permutations);
std::cout << "Number of permutations " << permutations.size() << "\n";
for(const std::string &s : permutations){
std::cout << s << "\n";
}
std::set <std::string> uniquePermutations(permutations.begin(), permutations.end());
std::cout << "Number of unique permutations " << uniquePermutations.size() << "\n";
for(const std::string &s : uniquePermutations){
std::cout << s << "\n";
}
return 0;
}
输出:
Number of permutations 6
112
121
112
121
211
211
Number of unique permutations 3
112
121
211
//Rotate and create words beginning with all letter possible and push to stack 1
//Read from stack1 and for each word create words with other letters at the next location by rotation and so on
/* eg : man
1. push1 - man, anm, nma
2. pop1 - nma , push2 - nam,nma
pop1 - anm , push2 - amn,anm
pop1 - man , push2 - mna,man
*/
public class StringPermute {
static String str;
static String word;
static int top1 = -1;
static int top2 = -1;
static String[] stringArray1;
static String[] stringArray2;
static int strlength = 0;
public static void main(String[] args) throws IOException {
System.out.println("Enter String : ");
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bfr = new BufferedReader(isr);
str = bfr.readLine();
word = str;
strlength = str.length();
int n = 1;
for (int i = 1; i <= strlength; i++) {
n = n * i;
}
stringArray1 = new String[n];
stringArray2 = new String[n];
push(word, 1);
doPermute();
display();
}
public static void push(String word, int x) {
if (x == 1)
stringArray1[++top1] = word;
else
stringArray2[++top2] = word;
}
public static String pop(int x) {
if (x == 1)
return stringArray1[top1--];
else
return stringArray2[top2--];
}
public static void doPermute() {
for (int j = strlength; j >= 2; j--)
popper(j);
}
public static void popper(int length) {
// pop from stack1 , rotate each word n times and push to stack 2
if (top1 > -1) {
while (top1 > -1) {
word = pop(1);
for (int j = 0; j < length; j++) {
rotate(length);
push(word, 2);
}
}
}
// pop from stack2 , rotate each word n times w.r.t position and push to
// stack 1
else {
while (top2 > -1) {
word = pop(2);
for (int j = 0; j < length; j++) {
rotate(length);
push(word, 1);
}
}
}
}
public static void rotate(int position) {
char[] charstring = new char[100];
for (int j = 0; j < word.length(); j++)
charstring[j] = word.charAt(j);
int startpos = strlength - position;
char temp = charstring[startpos];
for (int i = startpos; i < strlength - 1; i++) {
charstring[i] = charstring[i + 1];
}
charstring[strlength - 1] = temp;
word = new String(charstring).trim();
}
public static void display() {
int top;
if (top1 > -1) {
while (top1 > -1)
System.out.println(stringArray1[top1--]);
} else {
while (top2 > -1)
System.out.println(stringArray2[top2--]);
}
}
}
使用递归。
依次尝试每个字母作为第一个字母,然后使用递归调用找到剩余字母的所有排列。 基本情况是,当输入是空字符串时,唯一的排列就是空字符串。
我的实现基于Mark Byers上面的描述:
static Set<String> permutations(String str){
if (str.isEmpty()){
return Collections.singleton(str);
}else{
Set <String> set = new HashSet<>();
for (int i=0; i<str.length(); i++)
for (String s : permutations(str.substring(0, i) + str.substring(i+1)))
set.add(str.charAt(i) + s);
return set;
}
}
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