使用迭代的字符串排列

Question

我正在尝试查找给定字符串的排列，但我想使用迭代。我在网上找到的递归解决方案我确实理解它，但是将其转换为迭代解决方案确实行不通。下面我附上了我的代码。非常感谢您的帮助：

public static void combString(String s) {
    char[] a = new char[s.length()];
    //String temp = "";
    for(int i = 0; i < s.length(); i++) {
        a[i] = s.charAt(i);
    }
    for(int i = 0; i < s.length(); i++) {
        String temp = "" + a[i];    

        for(int j = 0; j < s.length();j++) {
            //int k = j;
            if(i != j) {
                System.out.println(j);
                temp += s.substring(0,j) + s.substring(j+1,s.length());
            }               
        }
        System.out.println(temp);
    }
}

Answer 1

跟进我的related question 评论，这是一个使用Counting QuickPerm Algorithm 完成您想要的Java 实现：

public static void combString(String s) {
    // Print initial string, as only the alterations will be printed later
    System.out.println(s);   
    char[] a = s.toCharArray();
    int n = a.length;
    int[] p = new int[n];  // Weight index control array initially all zeros. Of course, same size of the char array.
    int i = 1; //Upper bound index. i.e: if string is "abc" then index i could be at "c"
    while (i < n) {
        if (p[i] < i) { //if the weight index is bigger or the same it means that we have already switched between these i,j (one iteration before).
            int j = ((i % 2) == 0) ? 0 : p[i];//Lower bound index. i.e: if string is "abc" then j index will always be 0.
            swap(a, i, j);
            // Print current
            System.out.println(join(a));
            p[i]++; //Adding 1 to the specific weight that relates to the char array.
            i = 1; //if i was 2 (for example), after the swap we now need to swap for i=1
        }
        else { 
            p[i] = 0;//Weight index will be zero because one iteration before, it was 1 (for example) to indicate that char array a[i] swapped.
            i++;//i index will have the option to go forward in the char array for "longer swaps"
        }
    }
}

private static String join(char[] a) {
    StringBuilder builder = new StringBuilder();
    builder.append(a);
    return builder.toString();
}

private static void swap(char[] a, int i, int j) {
    char temp = a[i];
    a[i] = a[j];
    a[j] = temp;
}

Answer 2

    List<String> results = new ArrayList<String>();
    String test_str = "abcd";
    char[] chars = test_str.toCharArray();
    results.add(new String("" + chars[0]));
    for(int j=1; j<chars.length; j++) {
        char c = chars[j];
        int cur_size = results.size();
        //create new permutations combing char 'c' with each of the existing permutations
        for(int i=cur_size-1; i>=0; i--) {
            String str = results.remove(i);
            for(int l=0; l<=str.length(); l++) {
                results.add(str.substring(0,l) + c + str.substring(l));
            }
        }
    }
    System.out.println("Number of Permutations: " + results.size());
    System.out.println(results);

示例： 如果我们有 3 个字符串，例如“abc”，我们可以形成如下排列。

1) 构造一个带有第一个字符的字符串，例如'a' 并将其存储在结果中。

    char[] chars = test_str.toCharArray();
    results.add(new String("" + chars[0]));

2) 现在取字符串中的下一个字符（即“b”）并将其插入结果中先前构造的字符串的所有可能位置。由于此时我们在结果中只有一个字符串（“a”），因此这样做会为我们提供 2 个新字符串 'ba'、'ab'。在结果中插入这些新构造的字符串并删除“a”。

    for(int i=cur_size-1; i>=0; i--) {
        String str = results.remove(i);
        for(int l=0; l<=str.length(); l++) {
            results.add(str.substring(0,l) + c + str.substring(l));
        }
    }

3) 对给定字符串中的每个字符重复 2)。

for(int j=1; j<chars.length; j++) {
    char c = chars[j];
     ....
     ....
}

这给了我们来自“ba”和“cab”的“cba”、“bca”、“bac”、来自“ab”的“acb”和“abc”

Answer 3

工作队列允许我们为这个问题创建一个优雅的迭代解决方案。

static List<String> permutations(String string) {
    List<String> permutations = new LinkedList<>();
    Deque<WorkUnit> workQueue = new LinkedList<>(); 

    // We need to permutate the whole string and haven't done anything yet.
    workQueue.add(new WorkUnit(string, ""));

    while (!workQueue.isEmpty()) { // Do we still have any work?
        WorkUnit work = workQueue.poll();

        // Permutate each character.
        for (int i = 0; i < work.todo.length(); i++) {
            String permutation = work.done + work.todo.charAt(i);

            // Did we already build a complete permutation?
            if (permutation.length() == string.length()) {
                permutations.add(permutation);
            } else {

                // Otherwise what characters are left? 
                String stillTodo = work.todo.substring(0, i) + work.todo.substring(i + 1);
                workQueue.add(new WorkUnit(stillTodo, permutation));
            }
        }
    }
    return permutations; 
}

保存部分结果的辅助类非常简单。

/**
 * Immutable unit of work
 */
class WorkUnit {
    final String todo;
    final String done;

    WorkUnit(String todo, String done) {
        this.todo = todo;
        this.done = done;
    }
}

您可以通过将上面的代码包装在这个类中来测试它们。

import java.util.*;

public class AllPermutations {

    public static void main(String... args) {
        String str = args[0];
        System.out.println(permutations(str));
    }

    static List<String> permutations(String string) {
        ...
    }
}

class WorkUnit {
    ...
}

编译运行试试。

$ javac AllPermutations.java; java AllPermutations abcd

通过使用 LIFO 工作堆栈而不是 FIFO 队列，还可以轻松地调整以下实现以以相反的顺序返回排列列表。

Answer 4

import java.util.List;
import java.util.Set;
import java.util.ArrayList;
import java.util.HashSet;

public class Anagrams{

    public static void main(String[] args)
    {

        String inpString = "abcd";
        Set<String> combs = getAllCombs(inpString);

        for(String comb : combs)
        {
            System.out.println(comb);
        }

    }


    private static Set<String> getAllCombs(String inpString)
    {
        Set<String> combs = new HashSet<String>();
        if( inpString == null | inpString.isEmpty())
            return combs;

        combs.add(inpString.substring(0,1));
        Set<String> tempCombs = new HashSet<String>();
        for(char a : inpString.substring(1).toCharArray())
        {
            tempCombs.clear();
            tempCombs.addAll(combs);
            combs.clear();
            for(String comb : tempCombs)
            {
                combs.addAll(getCombs(comb,a));
            }
        }
        return combs;
    }

    private static Set<String> getCombs(String comb, char a) {
        Set<String> combs = new HashSet<String>();
        for(int i = 0 ; i <= comb.length(); i++)
        {
            String temp = comb.substring(0, i) + a + comb.substring(i);
            combs.add(temp);
            //System.out.println(temp);
        }
        return combs;
    }   
}

Answer 5

只是发布我解决问题的方法：

import java.util.ArrayDeque;
import java.util.Queue;

public class PermutationIterative {
    public static void main(String[] args) {
        permutationIterative("abcd");
    }

    private static void permutationIterative(String str) {
        Queue<String> currentQueue = null;
        int charNumber = 1;
        for (char c : str.toCharArray()) {
            if (currentQueue == null) {
                currentQueue = new ArrayDeque<>(1);
                currentQueue.add(String.valueOf(c));
            } else {
                int currentQueueSize = currentQueue.size();
                int numElements = currentQueueSize * charNumber;
                Queue<String> nextQueue = new ArrayDeque<>(numElements);
                for (int i = 0; i < currentQueueSize; i++) {
                    String tempString = currentQueue.remove();
                    for (int j = 0; j < charNumber; j++) {
                        int n = tempString.length();
                        nextQueue.add(tempString.substring(0, j) + c + tempString.substring(j, n));
                    }
                }
                currentQueue = nextQueue;
            }
            charNumber++;
        }
        System.out.println(currentQueue);
    }
}

Answer 6

package vishal villa;

import java.util.Scanner;

public class Permutation {
    static void result( String st, String ans)
    {

        if(st.length() == 0)
            System.out.println(ans +" ");

        for(int i = 0; i<st.length(); i++)
        {
            char ch = st.charAt(i);
            String r = st.substring(0, i) +  st.substring(i + 1);
            result(r, ans + ch);    
        }   
    }
    public static void main(String[] args) 
    {
        Scanner Sc = new Scanner(System.in);
        System.out.println("enter the string");
        String st = Sc.nextLine();
    Permutation p = new Permutation();
        p.result(st,"" );
    }
}

Answer 7

// Java program to print all permutations of a
// given string.
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 s starting index
     * @param e end index
     */
    private void permute(String str, int s, int e)
    {
        if (s == e)
            System.out.println(str);
        else
        {
            for (int i = s; i <= s; i++)
            {
                str = swap(str,l,i);
                permute(str, s+1, e);
                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);
    }

}