python reverse linked list
def reverseLinkedList(root):
cur = root
prev = None
while cur:
nxt = cur.next
cur.next = prev
prev = cur
cur = nxt
return prev
python reverse linked list
def reverseLinkedList(root):
cur = root
prev = None
while cur:
nxt = cur.next
cur.next = prev
prev = cur
cur = nxt
return prev
reverse a linked list
class recursion {
static Node head; // head of list
static class Node {
int data;
Node next;
Node(int d)
{ data = d;
next = null; } }
static Node reverse(Node head)
{
if (head == null || head.next == null)
return head;
/* reverse the rest list and put the first element
at the end */
Node rest = reverse(head.next);
head.next.next = head;
/* tricky step -- see the diagram */
head.next = null;
/* fix the head pointer */
return rest;
} /* Function to print linked list */
static void print()
{
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
static void push(int data)
{
Node temp = new Node(data);
temp.next = head;
head = temp;
} /* Driver program to test above function*/
public static void main(String args[])
{
/* Start with the empty list */
push(20);
push(4);
push(15);
push(85);
System.out.println("Given linked list");
print();
head = reverse(head);
System.out.println("Reversed Linked list");
print();
} } // This code is contributed by Prakhar Agarwal
linked list reverse
// using iterative method to reverse linked list in JavaScript
// time complexity: O(n) & space complexity: O(1)
reverse() {
if (!this.head.next) {
return this.head;
}
let prevNode = null;
let currNode = this.head;
let nextNode = this.head;
while(nextNode){
nextNode = currNode.next;
currNode.next = prevNode;
prevNode = currNode;
currNode = nextNode;
}
this.head = prevNode;
return this.printList();
}
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