问题
Why is the time complexity of node deletion in doubly linked lists (O(1)) faster than node deletion in singly linked lists (O(n))?
回答1:
The problem assumes that the node to be deleted is known and a pointer to that node is available.
In order to delete a node and connect the previous and the next node together, you need to know their pointers. In a doubly-linked list, both pointers are available in the node that is to be deleted. The time complexity is constant in this case, i.e., O(1).
Whereas in a singly-linked list, the pointer to the previous node is unknown and can be found only by traversing the list from head until it reaches the node that has a next node pointer to the node that is to be deleted. The time complexity in this case is O(n).
In cases where the node to be deleted is known only by value, the list has to be searched and the time complexity becomes O(n) in both singly- and doubly-linked lists.
回答2:
Actually deletion in singly linked lists can also be implemented in O(1).
Given a singly linked list with the following state:
SinglyLinkedList:
Node 1 -> Node 2
Node 2 -> Node 3
Node 3 -> Node 4
Node 4 -> None
Head = Node 1
We can implement delete Node 2 in such a way:
Node 2 Value <- Node 3 Value
Node 2 -> Node 4
Here we replace the value of Node 2 with the value of its next node (Node 3) and set its next value pointer to the next value pointer of Node 3 (Node 4), skipping over the now effectively "duplicate" Node 3. Thus no traversal needed.
回答3:
Because you can't look backwards...
回答4:
Unless the element to be deleted is the head(or first) node, we need to traverse to the node before the one to be deleted. Hence, in worst case, i.e., when we need to delete the last node, the pointer has to go all the way to the second last node thereby traversing (n-1) positions, which gives us a time complexity of O(n).
回答5:
It has to do with the complexity of fixing up the next pointer in the node previous to the one you're deleting.
回答6:
Insertion and deletion at a known position is O(1). However, finding that position is O(n), unless it is the head or tail of the list.
When we talk about insertion and deletion complexity, we generally assume we already know where that's going to occur.
回答7:
I don't think Its O(1) unless you know the address of the node whichh has to be deleted ..... Don't you loop to reach the node which has to be deleted from head ????
It is O(1) provided you have the address of the node which has to be deleted because you have it's prev node link and next node link . As you have all the necessary links available just make the "node of interest " out of the list by re arranging the links and then free() it .
But in a single linked list you have to traverse from head to get it's previous and next address doesn't matter whether you have the address to f the node to be deleted or the node position ( as in 1st ,2nd ,10th etc.,.) To be deleted .
来源：https://stackoverflow.com/questions/1895769/time-complexity-of-node-deletion-in-singly-and-doubly-linked-lists