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
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));
}
其他回答
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;
}
让我们以输入abc为例。
从集合(["c"])中的最后一个元素(c)开始,然后将最后第二个元素(b)添加到它的前面,末尾和中间的每个可能位置,使其["bc", "cb"],然后以同样的方式将后面的下一个元素(a)添加到集合中的每个字符串中,使其:
"a" + "bc" = ["abc", "bac", "bca"] and "a" + "cb" = ["acb" ,"cab", "cba"]
因此整个排列:
["abc", "bac", "bca","acb" ,"cab", "cba"]
代码:
public class Test
{
static Set<String> permutations;
static Set<String> result = new HashSet<String>();
public static Set<String> permutation(String string) {
permutations = new HashSet<String>();
int n = string.length();
for (int i = n - 1; i >= 0; i--)
{
shuffle(string.charAt(i));
}
return permutations;
}
private static void shuffle(char c) {
if (permutations.size() == 0) {
permutations.add(String.valueOf(c));
} else {
Iterator<String> it = permutations.iterator();
for (int i = 0; i < permutations.size(); i++) {
String temp1;
for (; it.hasNext();) {
temp1 = it.next();
for (int k = 0; k < temp1.length() + 1; k += 1) {
StringBuilder sb = new StringBuilder(temp1);
sb.insert(k, c);
result.add(sb.toString());
}
}
}
permutations = result;
//'result' has to be refreshed so that in next run it doesn't contain stale values.
result = new HashSet<String>();
}
}
public static void main(String[] args) {
Set<String> result = permutation("abc");
System.out.println("\nThere are total of " + result.size() + " permutations:");
Iterator<String> it = result.iterator();
while (it.hasNext()) {
System.out.println(it.next());
}
}
}
我定义了左右两个字符串。一开始,左边是输入字符串,右边是“”。我递归地从左边选择所有可能的字符,并将其添加到右边的末尾。然后,在left-charAt(I)和right+charAt(I)上调用递归函数。我定义了一个类来跟踪生成的排列。
import java.util.HashSet;
import java.util.Set;
public class FindPermutations {
static class Permutations {
Set<String> permutations = new HashSet<>();
}
/**
* Building all the permutations by adding chars of left to right one by one.
*
* @param left The left string
* @param right The right string
* @param permutations The permutations
*/
private void findPermutations(String left, String right, Permutations permutations) {
int n = left.length();
if (n == 0) {
permutations.permutations.add(right);
}
for (int i = 0; i < n; i++) {
findPermutations(left.substring(0, i) + left.substring(i + 1, n), right + left.charAt(i), permutations);
}
}
/**
* Gets all the permutations of a string s.
*
* @param s The input string
* @return all the permutations of a string s
*/
public Permutations getPermutations(String s) {
Permutations permutations = new Permutations();
findPermutations(s, "", permutations);
return permutations;
}
public static void main(String[] args) {
FindPermutations findPermutations = new FindPermutations();
String s = "ABC";
Permutations permutations = findPermutations.getPermutations(s);
printPermutations(permutations);
}
private static void printPermutations(Permutations permutations) {
for (String p : permutations.permutations) {
System.out.println(p);
}
}
}
我希望这能有所帮助。
Java中一个非常基本的解决方案是使用递归+设置(以避免重复),如果你想存储和返回解决方案字符串:
public static Set<String> generatePerm(String input)
{
Set<String> set = new HashSet<String>();
if (input == "")
return set;
Character a = input.charAt(0);
if (input.length() > 1)
{
input = input.substring(1);
Set<String> permSet = generatePerm(input);
for (String x : permSet)
{
for (int i = 0; i <= x.length(); i++)
{
set.add(x.substring(0, i) + a + x.substring(i));
}
}
}
else
{
set.add(a + "");
}
return set;
}
我们可以用阶乘来计算有多少字符串以某个字母开头。
示例:取输入abcd。(3!) == 6个字符串将以abcd中的每个字母开头。
static public int facts(int x){
int sum = 1;
for (int i = 1; i < x; i++) {
sum *= (i+1);
}
return sum;
}
public static void permutation(String str) {
char[] str2 = str.toCharArray();
int n = str2.length;
int permutation = 0;
if (n == 1) {
System.out.println(str2[0]);
} else if (n == 2) {
System.out.println(str2[0] + "" + str2[1]);
System.out.println(str2[1] + "" + str2[0]);
} else {
for (int i = 0; i < n; i++) {
if (true) {
char[] str3 = str.toCharArray();
char temp = str3[i];
str3[i] = str3[0];
str3[0] = temp;
str2 = str3;
}
for (int j = 1, count = 0; count < facts(n-1); j++, count++) {
if (j != n-1) {
char temp1 = str2[j+1];
str2[j+1] = str2[j];
str2[j] = temp1;
} else {
char temp1 = str2[n-1];
str2[n-1] = str2[1];
str2[1] = temp1;
j = 1;
} // end of else block
permutation++;
System.out.print("permutation " + permutation + " is -> ");
for (int k = 0; k < n; k++) {
System.out.print(str2[k]);
} // end of loop k
System.out.println();
} // end of loop j
} // end of loop i
}
}
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