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
这是一个C解:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
char* addLetter(char* string, char *c) {
char* result = malloc(sizeof(string) + 2);
strcpy(result, string);
strncat(result, c, 1);
return result;
}
char* removeLetter(char* string, char *c) {
char* result = malloc(sizeof(string));
int j = 0;
for (int i = 0; i < strlen(string); i++) {
if (string[i] != *c) {
result[j++] = string[i];
}
}
result[j] = '\0';
return result;
}
void makeAnagram(char *anagram, char *letters) {
if (*letters == '\0') {
printf("%s\n", anagram);
return;
}
char *c = letters;
while (*c != '\0') {
makeAnagram(addLetter(anagram, c),
removeLetter(letters, c));
c++;
}
}
int main() {
makeAnagram("", "computer");
return 0;
}
其他回答
为排列和组合添加更详细的NcK/NcR
public static void combinationNcK(List<String> inputList, String prefix, int chooseCount, List<String> resultList) {
if (chooseCount == 0)
resultList.add(prefix);
else {
for (int i = 0; i < inputList.size(); i++)
combinationNcK(inputList.subList(i + 1, inputList.size()), prefix + "," + inputList.get(i), chooseCount - 1, resultList);
// Finally print once all combinations are done
if (prefix.equalsIgnoreCase("")) {
resultList.stream().map(str -> str.substring(1)).forEach(System.out::println);
}
}
}
public static void permNcK(List<String> inputList, int chooseCount, List<String> resultList) {
for (int count = 0; count < inputList.size(); count++) {
permNcK(inputList, "", chooseCount, resultList);
resultList = new ArrayList<String>();
Collections.rotate(inputList, 1);
System.out.println("-------------------------");
}
}
public static void permNcK(List<String> inputList, String prefix, int chooseCount, List<String> resultList) {
if (chooseCount == 0)
resultList.add(prefix);
else {
for (int i = 0; i < inputList.size(); i++)
combinationNcK(inputList.subList(i + 1, inputList.size()), prefix + "," + inputList.get(i), chooseCount - 1, resultList);
// Finally print once all combinations are done
if (prefix.equalsIgnoreCase("")) {
resultList.stream().map(str -> str.substring(1)).forEach(System.out::println);
}
}
}
public static void main(String[] args) {
List<String> positions = Arrays.asList(new String[] { "1", "2", "3", "4", "5", "6", "7", "8" });
List<String> resultList = new ArrayList<String>();
//combinationNcK(positions, "", 3, resultList);
permNcK(positions, 3, resultList);
}
//插入每个字符到数组列表中
static ArrayList al = new ArrayList();
private static void findPermutation (String str){
for (int k = 0; k < str.length(); k++) {
addOneChar(str.charAt(k));
}
}
//insert one char into ArrayList
private static void addOneChar(char ch){
String lastPerStr;
String tempStr;
ArrayList locAl = new ArrayList();
for (int i = 0; i < al.size(); i ++ ){
lastPerStr = al.get(i).toString();
//System.out.println("lastPerStr: " + lastPerStr);
for (int j = 0; j <= lastPerStr.length(); j++) {
tempStr = lastPerStr.substring(0,j) + ch +
lastPerStr.substring(j, lastPerStr.length());
locAl.add(tempStr);
//System.out.println("tempStr: " + tempStr);
}
}
if(al.isEmpty()){
al.add(ch);
} else {
al.clear();
al = locAl;
}
}
private static void printArrayList(ArrayList al){
for (int i = 0; i < al.size(); i++) {
System.out.print(al.get(i) + " ");
}
}
作为Python生成器,带有现代类型提示:
from typing import Iterator
def permutations(string: str, prefix: str = '') -> Iterator[str]:
if len(string) == 0:
yield prefix
for i, character in enumerate(string):
yield from permutations(string[:i] + string[i + 1:], prefix + character)
for p in permutations('abcd'):
print(p)
如果有人想要生成排列来做一些事情,而不是通过void方法打印它们:
static List<int[]> permutations(int n) {
class Perm {
private final List<int[]> permutations = new ArrayList<>();
private void perm(int[] array, int step) {
if (step == 1) permutations.add(array.clone());
else for (int i = 0; i < step; i++) {
perm(array, step - 1);
int j = (step % 2 == 0) ? i : 0;
swap(array, step - 1, j);
}
}
private void swap(int[] array, int i, int j) {
int buffer = array[i];
array[i] = array[j];
array[j] = buffer;
}
}
int[] nVector = new int[n];
for (int i = 0; i < n; i++) nVector [i] = i;
Perm perm = new Perm();
perm.perm(nVector, n);
return perm.permutations;
}
这对我很有效。
import java.util.Arrays;
public class StringPermutations{
public static void main(String args[]) {
String inputString = "ABC";
permute(inputString.toCharArray(), 0, inputString.length()-1);
}
public static void permute(char[] ary, int startIndex, int endIndex) {
if(startIndex == endIndex){
System.out.println(String.valueOf(ary));
}else{
for(int i=startIndex;i<=endIndex;i++) {
swap(ary, startIndex, i );
permute(ary, startIndex+1, endIndex);
swap(ary, startIndex, i );
}
}
}
public static void swap(char[] ary, int x, int y) {
char temp = ary[x];
ary[x] = ary[y];
ary[y] = temp;
}
}
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