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
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。
其他回答
import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;
public class hello {
public static void main(String[] args) throws IOException {
hello h = new hello();
h.printcomp();
}
int fact=1;
public void factrec(int a,int k){
if(a>=k)
{fact=fact*k;
k++;
factrec(a,k);
}
else
{System.out.println("The string will have "+fact+" permutations");
}
}
public void printcomp(){
String str;
int k;
Scanner in = new Scanner(System.in);
System.out.println("enter the string whose permutations has to b found");
str=in.next();
k=str.length();
factrec(k,1);
String[] arr =new String[fact];
char[] array = str.toCharArray();
while(p<fact)
printcomprec(k,array,arr);
// if incase u need array containing all the permutation use this
//for(int d=0;d<fact;d++)
//System.out.println(arr[d]);
}
int y=1;
int p = 0;
int g=1;
int z = 0;
public void printcomprec(int k,char array[],String arr[]){
for (int l = 0; l < k; l++) {
for (int b=0;b<k-1;b++){
for (int i=1; i<k-g; i++) {
char temp;
String stri = "";
temp = array[i];
array[i] = array[i + g];
array[i + g] = temp;
for (int j = 0; j < k; j++)
stri += array[j];
arr[z] = stri;
System.out.println(arr[z] + " " + p++);
z++;
}
}
char temp;
temp=array[0];
array[0]=array[y];
array[y]=temp;
if (y >= k-1)
y=y-(k-1);
else
y++;
}
if (g >= k-1)
g=1;
else
g++;
}
}
基于Heap算法的我的实现:
import java.util.ArrayList;
import java.util.List;
public class PermutationString {
public static List<String> permute(char[] str, int n) {
List<String> permutations = new ArrayList<>();
if (n == 1) {
permutations.add(new String(str));
}
else {
for (int i = 0; i < n; i++) {
permutations.addAll(permute(str, n-1));
if (n % 2 == 0) {
swap(str, i, n-1);
}
else {
swap(str, 0, n-1);
}
}
}
return permutations;
}
public static void swap(char[] str, int i, int j) {
char temp = str[i];
str[i] = str[j];
str[j] = temp;
}
public static void main(String[] args) {
List<String> permutations = permute("abcdefgh".toCharArray(), 8);
System.out.println(permutations);
}
}
时间复杂度为O(n!* n), O(n)为空间复杂度。
这就是我通过对排列和递归函数调用的基本理解所做的。虽然要花点时间,但都是独立完成的。
public class LexicographicPermutations {
public static void main(String[] args) {
// TODO Auto-generated method stub
String s="abc";
List<String>combinations=new ArrayList<String>();
combinations=permutations(s);
Collections.sort(combinations);
System.out.println(combinations);
}
private static List<String> permutations(String s) {
// TODO Auto-generated method stub
List<String>combinations=new ArrayList<String>();
if(s.length()==1){
combinations.add(s);
}
else{
for(int i=0;i<s.length();i++){
List<String>temp=permutations(s.substring(0, i)+s.substring(i+1));
for (String string : temp) {
combinations.add(s.charAt(i)+string);
}
}
}
return combinations;
}}
生成输出为[abc, acb, bac, bca, cab, cba]。
它背后的基本逻辑是
对于每个字符,将其视为第一个字符,并找出剩余字符的组合。例[abc](abc的组合)->。
a->[bc](a x Combination of (bc))->{abc,acb} b->[ac](b x组合(ac))->{bac,bca} c->[ab](c x Combination of (ab))->{cab,cba}
然后递归地分别调用每个[bc],[ac]和[ab]。
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;
}
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));
}
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