Question

我在python中编写了一些非常好用的代码，我试图用Java实现它，我遇到了一些麻烦。这是一个简单的21问题，使用回溯来获取用户输入的字符串，如＆＃34; 1 9 10＆＃34;

这是python代码：

def twentyone(nums, stack = [], answer = set()):
    for index, num in enumerate(nums):
        new_stack = stack + [num]
        total = sum(new_stack)
        if total == 21:
            answer.add(tuple(new_stack))
        elif total < 21:
            twentyone(nums[index + 1:], new_stack, answer)
    return answer

user_input = input()
list_format = [int(x) for x in user_input.split()]
answer = twentyone(list_format)

if len(answer) == 0:
    print("No combination of numbers add to 21")
for solution in answer:
    print("The values ", end = "")
    for number in solution:
            print("{} ".format(number), end = "")
    print("add up to 21")

这是我的java代码（到目前为止）

public class TwentyOne {

   public static void main(String args[]){
      Scanner userInput = new Scanner(System.in);
      String stringInput = userInput.nextLine();

      ArrayList<Integer> allNum = new ArrayList<>();
      for (String num : stringInput.split(" ")){
         allNum.add(Integer.parseInt(num));
      }

      HashSet<ArrayList<Integer>> answer = twentyOne(allNum);

      for (ArrayList<Integer> solution : answer){
         System.out.print("The values ");
         for (Integer num : solution){
            System.out.print(num + " ");
         }
         System.out.println("Add up to 21");
      }
   }



   private static HashSet twentyOne(ArrayList<Integer> user_input){

      return new HashSet<ArrayList<Integer>>();
   }
}

基本上，我不能在参数中没有变量初始化的情况下编写java递归问题。

即def twentyone(nums, stack = [], answer = set()):

所以我的问题是，如何在没有变量初始化的情况下处理java回溯，我将进行递归调用？

Answer 1

您将编写一个传递默认参数的包装器方法。 e.g：

twentyone(List<Integer> nums) {
    twentyoneRecursive(nums, new ArrayDeque<>(), new HashSet<>());
}

twentyoneRecursive(List<Integer> nums, Deque<Integer> stack, Set<List<Integer>> answer) {
    ...
    twentyoneRecursive(...);
}

Java代码转换的Python代码

1 个答案: