问题
One method which I can think of is to reverse the list and then read it.
But this involves changing the list which is bad.
OR I can make a copy of the list and then reverse it, but this uses additional O(n) memory.
Is there any better method which doesn't use extra memory and doesn't modify the list and runs in O(n) time
reverse linked list code is something like this in c#
Void Reverse (Node head)
{
Node prev= null;
Node current = head;
Node nextNode = null;
while (current!=null)
{
nextNode = current.Next;
current.Next = prev;
prev=current;
current = nextNode;
}
head = prev;
}
Recursive solution is
void ReadBackWard (Node n)
{
if (n==null)
return;
else
ReadBackward(n.Next);
Console.WriteLine(n.Data);
}
回答1:
To use O(n) memory and O(n) performance, create a stack; push everything on as you iterate in the forwards direction, then pop everything off, yielding the results.
To use O(n^2) performance (but O(1) extra memory), read it forwards each time, up the the node before the last one you got to.
Example:
IEnumerable<T> Reverse (Node head) {
Stack<Node> nodes = new Stack<Node>();
while(head != null) {
nodes.Push(head);
head = head.Next;
}
while(nodes.Count > 0) {
yield return nodes.Pop().Value;
}
}
回答2:
One of the hallmarks of a singly-linked list is that it is, in fact, singly linked. It is a one-way street, and there's no way to overcome that unless you turn it into something else (such as a reversed singly-linked list, a stack, a doubly-linked list...). One must be true to the nature of things.
As has been pointed out earlier; if you need to traverse a list both ways; you need to have a doubly-linked list. That is the nature of a doubly linked list, it goes both ways.
回答3:
Really you should be using a doubly-linked list.
If this isn't possible, I think your best option will be to construct a copy of the list that has been reversed.
Other options, such as relying on recursion (effectively copying the list to the stack) could cause you to run out of stack space if the list is too long.
回答4:
If you short of memory you can reverse list, iterate over it and reverse it again. Alternatively you can make a stack of pointers to nodes (or whatever is like a pointer in C#).
回答5:
void reverse_print(node *head)
{
node *newHead = NULL, *cur = head;
if(!head) return;
// Reverse the link list O(n) time O(1) space
while(cur){
head = head->next;
cur->next = newHead;
newHead = cur;
cur = head;
}
// Print the list O(n) time O(1) space
cur = newHead;
while(cur) {
printf(" %d", cur->val);
cur = cur->next;
}
// Reverse the link list again O(n) time O(1) space
cur = newHead;
while(cur){
newHead = newHead->next;
cur->next = head;
head = cur;
cur = newHead;
}
// Total complexity O(n) time O(1) space
}
回答6:
There is a third solution, this time using O(log(n)) memory and O(n log(n)) time, thus occupying the middle ground between the two solutions in Marc's answer.
It is effectively a reverse in-order descent of a binary tree [O(log(n))], except at each step you need to find the top of the tree [O(n)]:
- Split the list in two
- Recurse into the second half of the list
- Print the value at the midpoint
- Recurse into the first half
Here is the solution in Python (I don't know C#):
def findMidpoint(head, tail):
pos, mid = head, head
while pos is not tail and pos.next is not tail:
pos, mid = pos.next.next, mid.next
return mid
def printReversed(head, tail=None):
if head is not tail:
mid = findMidpoint(head, tail)
printReversed(mid.next, tail)
print mid.value,
printReversed(head, mid)
This could be recast using iteration instead of recursion, but at the cost of clarity.
For example, for a million-entry list, the three solutions take on the order of:
Solution Memory Performance ========================================= Marc #1 4MB 1 million operations Mine 80B 20 million operations Marc #2 4B 1 trillion operations
回答7:
Assuming your singly-linked list implements IEnumerable<T>, you can utilize LINQ's Reverse extension method:
var backwards = singlyLinkedList.Reverse();
You'll need to add a using System.Linq; directive at the top of the code file to use LINQ's extension methods.
回答8:
A variation of creating a stack and pushing all the elements onto the stack is to use recursion (and the system's built in stack), this is probably not the way to go with production code but serves as a better (IMHO) interview answer for the following reasons:
- It shows that you grok recursion
- It's less code and appears more elegant
- A naive interviewer may not realize that there is a space overhead (if this is the case you may want to consider whether you want to work there).
回答9:
Well, the naive solution would be to keep track of which node you're currently at, then iterate from the start until you find that node, always saving the node you just left. Then each time you find the node you're currently at, you produce the node you just left, save that node as the one you're currently at, then re-iterate from the start.
This would of course be horribly bad performance-wise.
I'm sure some smarter people have a better solution.
Pseudo-code (with bugs even):
current node = nothing
while current node is not first node
node = start
while node is not current node
previous node = node
node = next node
produce previous node
set current node to previous node
回答10:
This is messy but works:
class SinglyLinkedList {
SinglyLinkedList next;
int pos;
SinglyLinkedList(int pos) {
this.pos = pos;
}
SinglyLinkedList previous(SinglyLinkedList startNode) {
if (startNode == this) return null;
if (startNode.next == this) return startNode;
else return previous(startNode.next);
}
static int count = 0;
static SinglyLinkedList list;
static SinglyLinkedList head;
static SinglyLinkedList tail;
public static void main (String [] args) {
init();
System.out.println("Head: " + head.pos);
System.out.println("Tail: " + tail.pos);
list = head;
System.out.print("List forwards: ");
while (list != null) {
System.out.print(list.pos + ",");
list = list.next;
}
list = tail;
System.out.print("\nList backwards: ");
while (list.previous(head) != null) {
System.out.print(list.pos + ",");
list = list.previous(head);
}
}
static void init() {
list = new SinglyLinkedList(0);
head = list;
while (count < 100) {
list.next = new SinglyLinkedList(++count);
list = list.next;
}
tail = list;
}
}
回答11:
If in the Explicit Stack program, we create a stack for just the data of each node (instead of creating the Stack of type <Node>, we create Stack of type <T>) wouldn't it be even better? Because we don't need to store any other information of the Node then.
IEnumerable<T> Reverse (Node<T> head) {
Stack<T> nodes = new Stack<T>();
while(head != null) {
nodes.Push(head.data);
head = head.Next;
}
while(nodes.Count > 0) {
yield return nodes.Pop();
}
}
回答12:
you could read it in O(n^2) -- every time go to the last node read and print out the previous one
回答13:
What's wrong with:
public void printBackwards(LinkedList sl){
ListIterator<Element> litr = sl.listIterator(sl.size());
Element temp;
while(litr.previousIndex() >= 0){
temp = litr.previous();
System.out.println(temp);
}
}
O(n) performance, O(1) memory and simple as do-re-mi!
来源：https://stackoverflow.com/questions/1116720/how-to-read-a-singly-linked-list-backwards