Question

我有以下递归方法，为给定字符串生成排列。我正在尝试为arraylist中生成的字符串创建一个返回类型，更具体地说，我正在尝试在jsp页面中打印输出。

 public static void permutation(String str) { 
      permutation("", str); 
 }

private static void permutation(String prefix, String str) {
     int n = str.length();
     if (n == 0) System.out.println(prefix);
     else {
          for (int i = 0; i < n; i++)
        permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+ n));
        }
}

Answer 1

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Permute {

    public static void main(String[] args) {
        System.out.println(findAllPermutations("abc"));
    }

    public static List<String> findAllPermutations(String s) {
        if (s == null) {
            throw new NullPointerException();
        }
        if (s.length() <= 1) {
            return Arrays.asList(s);
        }

        List<String> permutations = new ArrayList<>();
        for (String permutation : findAllPermutations(s.substring(1))) {
            char ch = s.charAt(0);
            for (int i = 0; i <= permutation.length(); i++) {
                String prefix = permutation.substring(0, i);
                String suffix = permutation.substring(i);
                permutations.add(prefix + ch + suffix);
            }
        }
        return permutations;
    }
}

Answer 2

您可以使用静态列表“收集”结果：

static List<String> perms = new ArrayList<>();

public static void main(String[] args) throws IOException {
    permutation("abc");
    System.out.println(Arrays.toString(perms.toArray())); // prints [abc, acb, bac, bca, cab, cba]
}

public static void permutation(String str) {
    permutation("", str);
}

private static void permutation(String prefix, String str) {
    int n = str.length();
    if (n == 0) {
        perms.add(prefix);
    }
    else {
        for (int i = 0; i < n; i++)
            permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1)); // you had a bug in the last index on this line
    }
}

或者，您可以创建列表并将其作为参数传递：

public static void main(String[] args) throws IOException {
    List<String> perms = new ArrayList<>();
    permutation("abc", perms);
    System.out.println(Arrays.toString(perms.toArray())); // print [abc, acb, bac, bca, cab, cba]
}

public static void permutation(String str, List<String> perms) {
    permutation("", str, perms);
}

private static void permutation(String prefix, String str, List<String> perms) {
    int n = str.length();
    if (n == 0) {
        perms.add(prefix);
    }
    else {
        for (int i = 0; i < n; i++)
            permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1), perms);
    }
}

创建递归方法的返回类型

2 个答案: