Question

嗨，这是我在求职面试中遇到的一个基本问题，我试图使用Java获取输入字符串的所有排列，遗憾的是我无法使其工作。

 import java.util.Scanner;
 public class Test2 {

static void permute(char[] x, int y){
    if (y == x.length)
    {
        for(int i = 0; i < x.length; i++){
            System.out.print(x[y]);
        }

    }
    else {
        for (int i = y; i < x.length;i++)
        {
            char temp = x[y];
            x[y] = x[i];
            x[i] = temp;

            permute(x, y + 1);

            temp = x[y];
            x[y] = x[i];
            x[i] = temp;
        }
    }
}
    public static void main(String [] Args){
        Scanner scan = new Scanner (System.in);
        System.out.println("Input any word :");
        String word = scan.nextLine();

        int n = word.length();
        char [] sequence = new char[n];
        for (int i = 0; i < n ; i++)
            sequence[i] = scan.next().charAt(0);
        System.out.println("These are the permutations: ");
            permute(sequence,0);
            }
 }

Answer 1

我正在给一个关于C的程序，我希望将它转换成您选择的语言会变得非常困难。

来源： - http://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/

Answer 2

/**
 * Generates all permutations of a string given that all characters of that string are different.
 *
 * @param s
 *     the input string
 *
 * @return a set of all permutations from the given string
 */
public Set<String> generatePermutations(String s) {
    Set<String> permutations = new HashSet<String>();
    //Handling error scenarios
    if (s == null) {
        return null;
    } else if (s.length() == 0) {
        permutations.add("");
        return permutations;
    }

    char firstCharacter = s.charAt(0); // first character
    String remaining = s.substring(1); // Full string without first character
    Set<String> words = generatePermutations(remaining);
    for (String word : words) {
        for (int i = 0; i <= word.length(); i++) {
            permutations.add(insertCharacter(word, firstCharacter, i));
        }
    }
    return permutations;
}

/**
 * Given a collection of numbers that might contain duplicates, return all possible unique permutations. For
 * example, [1,1,2] have the following unique permutations: [1,1,2], [1,2,1], and [2,1,1].
 *
 * @param numbers
 *     the collection of integer numbers
 *
 * @return a set of all unique combinations from the given collection of numbers
 */
public HashSet<List<Integer>> generateCombinations(List<Integer> numbers) {
    Preconditions.checkNotNull(numbers);

    HashSet<List<Integer>> combinations = new HashSet<List<Integer>>();

    if (numbers.size() == 0) {
        combinations.add(new ArrayList<Integer>());
        return combinations;
    }

    int size = numbers.size();
    int lastNumber = numbers.get(size - 1);
    numbers.remove(size - 1);
    HashSet<List<Integer>> elements = generateCombinations(numbers);

    for (List<Integer> element : elements) {

        for (int i = 0; i < size; i++) {
            List<Integer> temp = new ArrayList<>(element);
            temp.add(i, lastNumber);
            // two Lists of the same base type (e.g. ArrayList) with the same content, in the same order, will compare as equal.
            //              // http://stackoverflow.com/questions/16657905/adding-arrays-with-same-values-to-hashset-results-in-duplicate-items
            combinations.add(temp);
        }
    }

    return combinations;
}

输入字符串的排列

2 个答案: