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
/*
* eg: abc =>{a,bc},{b,ac},{c,ab}
* =>{ca,b},{cb,a}
* =>cba,cab
* =>{ba,c},{bc,a}
* =>bca,bac
* =>{ab,c},{ac,b}
* =>acb,abc
*/
public void nonRecpermute(String prefix, String word)
{
String[] currentstr ={prefix,word};
Stack<String[]> stack = new Stack<String[]>();
stack.add(currentstr);
while(!stack.isEmpty())
{
currentstr = stack.pop();
String currentPrefix = currentstr[0];
String currentWord = currentstr[1];
if(currentWord.equals(""))
{
System.out.println("Word ="+currentPrefix);
}
for(int i=0;i<currentWord.length();i++)
{
String[] newstr = new String[2];
newstr[0]=currentPrefix + String.valueOf(currentWord.charAt(i));
newstr[1] = currentWord.substring(0, i);
if(i<currentWord.length()-1)
{
newstr[1] = newstr[1]+currentWord.substring(i+1);
}
stack.push(newstr);
}
}
}
其他回答
//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Set操作建模“依赖于其他选择的选择”更容易理解相关排列 使用依赖排列,可用的选择减少,因为位置被从左到右的选定字符填充。递归调用的终端条件是测试可用选择集是否为空。当满足终端条件时,置换完成,并存储到“结果”列表中。
public static List<String> stringPermutation(String s) {
List<String> results = new ArrayList<>();
Set<Character> charSet = s.chars().mapToObj(m -> (char) m).collect(Collectors.toSet());
stringPermutation(charSet, "", results);
return results;
}
private static void stringPermutation(Set<Character> charSet,
String prefix, List<String> results) {
if (charSet.isEmpty()) {
results.add(prefix);
return;
}
for (Character c : charSet) {
Set<Character> newSet = new HashSet<>(charSet);
newSet.remove(c);
stringPermutation(newSet, prefix + c, results);
}
}
该代码可以泛化为一组对象查找排列。在本例中,我使用了一组颜色。
public enum Color{
ORANGE,RED,BULE,GREEN,YELLOW;
}
public static List<List<Color>> colorPermutation(Set<Color> colors) {
List<List<Color>> results = new ArrayList<>();
List<Color> prefix = new ArrayList<>();
permutation(colors, prefix, results);
return results;
}
private static <T> void permutation(Set<T> set, List<T> prefix, List<List<T>> results) {
if (set.isEmpty()) {
results.add(prefix);
return;
}
for (T t : set) {
Set<T> newSet = new HashSet<>(set);
List<T> newPrefix = new ArrayList<>(prefix);
newSet.remove(t);
newPrefix.add(t);
permutation(newSet, newPrefix, results);
}
}
测试代码。
public static void main(String[] args) {
List<String> stringPerm = stringPermutation("abcde");
System.out.println("# of permutations:" + stringPerm.size());
stringPerm.stream().forEach(e -> System.out.println(e));
Set<Color> colorSet = Arrays.stream(Color.values()).collect(Collectors.toSet());
List<List<Color>> colorPerm = colorPermutation(colorSet);
System.out.println("# of permutations:" + colorPerm.size());
colorPerm.stream().forEach(e -> System.out.println(e));
}
public class StringPermutation {
// Function to print all the permutations of str
static void printPermutn(String str, String ans) {
// If string is empty
if (str.length() == 0) {
System.out.print(ans + " ");
return;
}
for (int i = 0; i < str.length(); i++) {
// ith character of str
char ch = str.charAt(i);
// Rest of the string after excluding
// the ith character
String ros = str.substring(0, i) + str.substring(i + 1);
// Recurvise call
printPermutn(ros, ans + ch);
}
}
public static void main(String[] args) {
String s = "ABC";
printPermutn(s, "");
}
}
/** Returns an array list containing all
* permutations of the characters in s. */
public static ArrayList<String> permute(String s) {
ArrayList<String> perms = new ArrayList<>();
int slen = s.length();
if (slen > 0) {
// Add the first character from s to the perms array list.
perms.add(Character.toString(s.charAt(0)));
// Repeat for all additional characters in s.
for (int i = 1; i < slen; ++i) {
// Get the next character from s.
char c = s.charAt(i);
// For each of the strings currently in perms do the following:
int size = perms.size();
for (int j = 0; j < size; ++j) {
// 1. remove the string
String p = perms.remove(0);
int plen = p.length();
// 2. Add plen + 1 new strings to perms. Each new string
// consists of the removed string with the character c
// inserted into it at a unique location.
for (int k = 0; k <= plen; ++k) {
perms.add(p.substring(0, k) + c + p.substring(k));
}
}
}
}
return perms;
}
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);
}
}
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