递归地从链接列表中删除节点

Question

Given the head of a linked list and the int to search for as parameters, I need a method that will remove the first occurrence of this number in the list, and return the modified list. however i cannot modify the original list. I know how to remove the node from the list, but im not sure how i would keep the original list intact since this has to be done recursively. below is the method ** initially M is the original list. I dont know that it will still be the same list after calling the method again...?

MyList removeNumber(MyList m, int removee){

Answer 1

//This function will always return a new list with 'remove' removed
MyList removeNumber(MyList m, int remove){
//if m is empty List, return an empty list

//if head is not the int to remove, return a New list from 
//   head concat removeNumber(m.next,remove)
//else return removeNumber(m.next,remove)
}

Answer 2

I think it lacks information. I'm assuming however a very traditional implementation for linked list, for instance:

class MyList {

  MyList prev;
  MyList next;
  int data;

  static MyList removeNumber(MyList m,int removee) {

    if(m == null) return null; // already empty

    if(m.data == removee) { // m already is the node to throw away

      if(m.prev != null)// relink
        m.prev.next = m.next;

      if(m.next != null)// relink
        m.next.prev = m.prev;

      return m.prev;
    }
    // if this node isn't the one yet, keep looking for
    return removeNumber(m.next,removee); 
  }
}

There are plenty different ways to do that, but you have to provide more info in order to allow us to point you the correct literature.

Answer 3

The idea is that the resulting structure will be a "Y": a two-headed list (actually a simple graph).

One branch of the Y is the original list. The other is your new list with removed node. The vertical stalk of the Y is what's after the element you remove. It's common to both lists. Here's some ascii art with the Y turned on its side showing a list of 1 to 5 with 3 removed.

     new -> 1 -> 2 ------\
                          v
original -> 1 -> 2 -> 3 -> 4 -> 5 -> null

Thinking recursively is all about defining a problem in terms of a smaller version of itself plus a fixed bit of work. And you need a base case (or maybe several).

A linked list is itself a recursive structure:

A list is either empty or it's an element linked by its "next" reference to a list.

Note this defines a list using a smaller list. The base case is the empty list. The fixed bit is the element.

Read this definition a few times, then see how it translates the code:

class MyList {
  int value;    // the element at the head of this list
  MyList next;  // the rest of the list

  MyList(int value, MyList next) {
    this.value = value;
    this.next = next;
  }
}

The base case "empty list" is just a null reference. The element removal problem expressed recursively using the same pattern becomes:

A copy of a list with an element removed is either a) the rest of the list following the head in the case that the element to be removed is the head or b) a copy of the current node followed by a copy the rest of the list with the desired element removed.

Here I'm defining a "copy of a list with one element removed" using a smaller version of the same thing. Case a) is the base case. The fixed bit is copying the head when it's not the removee .

Of course there's another base case: if the list is empty, the removee can't be found. That's an error.

Putting this in code:

MyList removeNumber(MyList m, int removee) {
  if (m == null) throw new RuntimeException("removee not found");
  if (m.value == removee) return m.next;
  return new MyList(m.value, removeNumber(m.next, removee));
}

Putting the function to use would look something like this:

MyList originalList = ... // list of 1 to 5.
MyList newListWith3removed = removeNumber(originalList, 3);

System.out.println("Original list:");
for (MyList p : originalList) System.out.println(p.value); 
System.out.println("With 3 removed:");
for (MyList p : newListWith3removed) System.out.println(p.value);

The output will look as expected: 1 to 5 in the first list and 1,2,4,5 in the second. Ie the first list is unchanged.