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Question

Is there a linear way to find the middle element of a singly linked list? (Linear - Meaning you can only iterate through the list once AKA the number of iterations you do total cannot exceed the length of the list)

Thanks!

Edit: The question specifies that you cannot know the length of the list beforehand.

Edit 2: The question is written for Java, but using some linked list definition that does not have a length() method

Answer 1

This might not fit your constraints, but they're quite vague. It only requires a single iteration of the list (in the sense of starting at the beginning only once, and reaching the end only once), but it requires two independent pointers being stored as you do so (and therefore follows half of the list's next pointers twice).

1. Set pointer_1 and pointer_2 to the first node in the list, and counter to 0
2. Set pointer_1 to pointer_1->next
3. Increment counter
4. If counter is even, set pointer_2 to pointer_2->next
5. If pointer_1 is not at the end of the list, goto 2
6. When the loop exits, pointer_2 is at the middle of the list

Answer 2

Use two variables that point to the first element of your list. Then iterate through the list, incrementing the first pointer by 1 place and the second pointer by 2 places on each iteration. Once the second pointer reaches the end of the list, the first pointer will contain the middle element (if it exists).

A bit of pseudo code...

list = [1, 2, 3, 4, 5]
ptr1 = list[0]
ptr2 = list[0]
while ptr2 is not null
    ptr1 = ptr1.next
    ptr2 = ptr2.next.next
return ptr1

Answer 3

Node ptr1 = head; 
Node ptr2 = head;

while(ptr1 != null || ptr1->next != null){
    ptr1 = ptr1 -> next -> next;
    ptr2 = ptr2 -> next; 
}

EDIT:

//this is wrong, so not needed so, upper part is enough. 
if(ptr1->next != null){
    ptr2 = ptr2 -> next;
}

Answer 4

Node p1 = head;
Node mid = head;

while(p1 != null){
    p1 = p1.next();
    p1 = p1.next();
    mid = mid.next(); 
}

Be aware, the order matters with p1 and mid in the loop. Take for instance if you had only one element in the list. That means the head would be the middle. However, mid would return null because it went to the next element, instead of exiting the loop before getting to the next element.

WRONG IMPLEMENTATION

while(p1 != null){
    mid = mid.next(); // incorrect order.
    p1 = p1.next().next(); //will cause exception if p1.next == null
}