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
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);
}
其他回答
import java.io.*;
public class Anagram {
public static void main(String[] args) {
java.util.Scanner sc=new java.util.Scanner(System.in);
PrintWriter p=new PrintWriter(System.out,true);
p.println("Enter Word");
String a[],s="",st;boolean flag=true;
int in[],n,nf=1,i,j=0,k,m=0;
char l[];
st=sc.next();
p.println("Anagrams");
p.println("1 . "+st);
l=st.toCharArray();
n=st.length();
for(i=1;i<=n;i++){
nf*=i;
}
i=1;
a=new String[nf];
in=new int[n];
a[0]=st;
while(i<nf){
for(m=0;m<n;m++){
in[m]=n;
}j=0;
while(j<n){
k=(int)(n*Math.random());
for(m=0;m<=j;m++){
if(k==in[m]){
flag=false;
break;
}
}
if(flag==true){
in[j++]=k;
}flag=true;
}s="";
for(j=0;j<n;j++){
s+=l[in[j]];
}
//Removing same words
for(m=0;m<=i;m++){
if(s.equalsIgnoreCase(a[m])){
flag=false;
break;
}
}
if(flag==true){
a[i++]=s;
p.println(i+" . "+a[i-1]);
}flag=true;
}
}
}
//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--]);
}
}
}
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);
}
}
倒计时Quickperm算法的通用实现,表示#1(可伸缩,非递归)。
/**
* Generate permutations based on the
* Countdown <a href="http://quickperm.org/">Quickperm algorithm</>.
*/
public static <T> List<List<T>> generatePermutations(List<T> list) {
List<T> in = new ArrayList<>(list);
List<List<T>> out = new ArrayList<>(factorial(list.size()));
int n = list.size();
int[] p = new int[n +1];
for (int i = 0; i < p.length; i ++) {
p[i] = i;
}
int i = 0;
while (i < n) {
p[i]--;
int j = 0;
if (i % 2 != 0) { // odd?
j = p[i];
}
// swap
T iTmp = in.get(i);
in.set(i, in.get(j));
in.set(j, iTmp);
i = 1;
while (p[i] == 0){
p[i] = i;
i++;
}
out.add(new ArrayList<>(in));
}
return out;
}
private static int factorial(int num) {
int count = num;
while (num != 1) {
count *= --num;
}
return count;
}
它需要list,因为泛型不能很好地使用数组。
这里有一个优雅的,非递归的O(n!)解:
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;
}
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