3

我希望这种方法适用于任何给定数量的参数，我可以通过代码生成来做到这一点（有很多丑陋的代码），它可以通过递归来完成吗？如果是这样怎么办？我理解递归，但我不知道如何写这个。

private static void allCombinations(List<String>... lists) {
    if (lists.length == 3) {
        for (String s3 : lists[0]) {
            for (String s1 : lists[1]) {
                for (String s2 : lists[2]) {
                    System.out.println(s1 + "-" + s2 + "-" + s3);
                }
            }
        }
    }
    if (lists.length == 2) {
        for (String s3 : lists[0]) {
            for (String s1 : lists[1]) {
                    System.out.println(s1 + "-" + s3);
            }
        }
    }
}
4

4 回答 4

3

这是一个简单的递归实现：

private static void allCombinations(List<String>... lists) {
  allCombinations(lists, 0, "");
}

private static void allCombinations(List<String>[] lists, int index, String pre) {
  for (String s : lists[index]) {
    if (index < lists.length - 1) {
      allCombinations(lists, index + 1, pre + s + "-");
    }else{
      System.out.println(pre + s);
    }
  }
}
于 2008-10-16T09:31:53.333 回答
1

你特别需要它是递归的吗？我会让它成为非递归的，但仍然不是特殊情况：

public static void allCombinations(List<String>... lists) {
    int[] indexes = new int[lists.length];

    while (incrementIndexes(lists, indexes)) {
        StringBuilder builder = new StringBuilder();
        for (int i=0; i < indexes.length; i++) {
            if (i != 0) {
                builder.append("-");
            }
            builder.append(lists[i].get(indexes[i]));
        }
        System.out.println(builder);
    }
}

private static boolean incrementIndexes(List<String>[] lists, int[] indexes) {
    for (int depth = indexes.length-1; depth >= 0; depth--) {
        indexes[depth]++;
        if (indexes[depth] != lists[depth].size()) {
            return true;
        }
        // Overflowed this index. Reset to 0 and backtrack
        indexes[depth] = 0;
    }
    // Everything is back to 0. Finished!
    return false;
}
于 2008-10-16T09:24:28.387 回答
1

这是一个广义的递归版本。它抱怨在测试代码中创建未经检查的通用数组，但置换代码本身是可以的：

import java.util.*;

public class Test
{
    public interface Action<T> {
        void execute(Iterable<T> values);
    }

    public static void main(String[] args) {
        List<String> first = Arrays.asList(new String[]{"1", "2", "3"});
        List<String> second = Arrays.asList(new String[]{"a", "b", "c"});
        List<String> third = Arrays.asList(new String[]{"x", "y"});
        Action<String> action = new Action<String>() {
            @Override public void execute(Iterable<String> values) {
                 StringBuilder builder = new StringBuilder();
                 for (String value : values) {
                     if (builder.length() != 0) {
                         builder.append("-");
                     }
                     builder.append(value);
                 }
                 System.out.println(builder);
            }
        };
        permute(action, first, second, third);
    }

    public static <T> void permute(Action<T> action, Iterable<T>... lists) {
        Stack<T> current = new Stack<T>();
        permute(action, lists, 0, current);
    }

    public static <T> void permute(Action<T> action, Iterable<T>[] lists,
        int index, Stack<T> current) {
        for (T element : lists[index]) {
            current.push(element);
            if (index == lists.length-1) {
              action.execute(current);
            } else {
              permute(action, lists, index+1, current);
            }
            current.pop();
        }
    }
}
于 2008-10-16T10:08:03.527 回答
0

这是我基于 Rasmus 解决方案的具有正确排序的递归解决方案。它仅在所有列表大小相同时才有效。

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


public class Test {

        public static void main(String[] args) {
        List<String> first = Arrays.asList(new String[]{"1", "2", "3"});
        List<String> second = Arrays.asList(new String[]{"a", "b", "c"});
        List<String> third = Arrays.asList(new String[]{"x", "y", "z"});
        allCombinations (first, second, third);
        }

        private static void allCombinations(List<String>... lists) {
              allCombinations(lists, 1, "");
        }

        private static void allCombinations(List<String>[] lists, int index, String pre) {
            int nextHop = hop(index, lists.length-1);
          for (String s : lists[index]) {
            if (index != 0) {
              allCombinations(lists, nextHop, pre + s + "-");
            } else System.out.println(pre + s);
          }
        }
        private static int hop(int prevIndex, int maxResult){
            if (prevIndex%2 == 0){
                return prevIndex-2;
            } else {
                if (prevIndex == maxResult) 
                    return prevIndex-1;
                int nextHop = prevIndex+2;
                if (nextHop > maxResult){
                    return maxResult;
                } else return nextHop;
            }
        }

}

允许不同大小列表的“正确排序”解决方案必须从最后一个列表开始并向后工作到第一个列表（lists [0]），将元素附加到“pre”字符串的开头或结尾并将其传递下去。同样，第一个列表将打印结果。我已经编码了，但是午餐已经准备好了，女朋友开始不喜欢 stackoverflow...

于 2008-10-16T10:50:19.727 回答