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我想根据用户输入找到任何大小的 2^n 排列。我不知道该怎么做。我知道我必须使用递归。与我见过的大多数示例不同，函数的参数只是数字输入，没有可用于进行所有排列的字符串。所有排列都将存储在 arrayList 中并输出给用户。即（输入3）输出：000 001 010 011 100 101 110 111 这是我到目前为止的代码：

public static void printBin(int bits) {
    ArrayList<String> binaryArray = new ArrayList<String>();

    if(bits == 0) {
        System.out.println(binaryArray);
    }
    else {
        for (String string2 : binaryArray) {
            String comp = "";
            if (comp.length() <= string2.length()) {
                comp = string2;
                string = comp;
            }
            string.concat("0");
            binaryArray.add(string);
            printBin(bits - 1);
            string.concat("1");
            binaryArray.add(string);
            printBin(bits-1);
        }


    }
}

提前致谢。

4

2 回答 2

1

好吧，我认为这是你能做的最好的。

import java.util.ArrayList;

public class BinaryList {

    public static void main(String[] args) {
        try {
            if (args.length != 1 || Integer.parseInt(args[0]) < 1)) {
                System.err.println("Invalid integer argument");
                return;
            }

            binaryRecursive(Integer.parseInt(args[0]), new ArrayList<String>(0));

        } catch (NumberFormatException e) {
            System.err.println("Argument not an integer");
        }
    }

    public static void binaryRecursive(int bits, ArrayList<String> list) {

        if (list.size() == (int)Math.pow(2, bits)) {
            for (String n : list) {
                System.out.println(n);
            }

        } else {
            StringBuilder n = new StringBuilder(bits);

            for (int i = bits - 1; i >= 0; i--) {
                n.append((list.size() >> i) & 1);
            }

            list.add(n.toString());

            binaryRecursive(bits, list);
        }
    }
}

如果不将最后一个作为参数传递、返回值或将列表保留为字段，则无法保留它们的列表。

按照 bit == 2 的逻辑，你得到的是：

* 1st method call

list.size() == 0

for 1 to 0 {
    (00 >> 1 = 00) & 01 == 0
    (00 >> 0 = 00) & 01 == 0
}

list.add("00")

* 2nd method call

list.size() == 1

for 1 to 0 {
    (01 >> 1 = 00) & 01 == 0
    (01 >> 0 = 01) & 01 == 1
}

list.add("01")

* 3rd method call

list.size() == 2

for 1 to 0 {
    (10 >> 1 = 01) & 01 == 1
    (10 >> 0 = 10) & 01 == 0
}

list.add("10")

* 4th method call

list.size() == 3

for 1 to 0 {
    (11 >> 1 = 01) & 01 == 1
    (11 >> 0 = 11) & 01 == 1
}

list.add("11")

* 5th method call

list.size() == 4 == 2 ^ 2

print the list
于 2013-10-28T01:14:06.833 回答
0

You start out with n.. I want n=3 bits. Your main function calls a function, "genBin", passing one arguments of ("0", n-1) and the other a ("1",n-1). genBin returns a List. The main function merges the two lists as one ArrayList.

genBin(String in, int bitsLeft), if bitsLeft > 0, calls itself with ("0"+in,bitsLeft-1) and ("1"+in,bitsLeft-1) and merges the two return values as an ArrayList. If bitsLeft == 0, it returns in as a single element ArrayList.

Your main function, upon doing that merge I previously mentioned, ends up with a merged List of all the permutations.

于 2013-10-28T01:04:31.817 回答