Leetcode - 206 - Reverse Linked List

Reverse a singly linked list.

Example:

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

 

Solutions:

(1) Iterative

We want the curr->next to be prev and in order to iterate in the loop, three elements(prev, curr, nextTemp) need to be updated each time. 

Java:

public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode curr = head;
    while (curr != null) {
        ListNode nextTemp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = nextTemp;
    }
    return prev;
}
// https://leetcode.com/problems/reverse-linked-list/solution/

C++:

ListNode* reverseList(ListNode* head) {
        ListNode *prev = NULL, *cur=head, *tmp;
        while(cur){
            tmp = cur->next;
            cur->next = prev;
            prev = cur;
            cur = tmp;
        }
        return prev;
  }
// by tamugaoqi from https://leetcode.com/problems/reverse-linked-list/discuss/58130/C%2B%2B-Iterative-and-Recursive

Python:

class Solution:
# @param {ListNode} head
# @return {ListNode}
def reverseList(self, head):
    prev = None
    while head:
        curr = head
        head = head.next
        curr.next = prev
        prev = curr
    return prev
// https://leetcode.com/problems/reverse-linked-list/discuss/58127/Python-Iterative-and-Recursive-Solution

 

(2) Recursive

Let's assume the list is: n1 → … → nk-1 → nk → nk+1 → … → nm → Ø

Assume from node nk+1 to nm had been reversed and you are at node nk.

n1 → … → nk-1 → nk → nk+1 ← … ← nm

We want nk+1’s next node to point to nk.

So,nk.next.next = nk;

n1's next must point to Ø. If you forget about this, your linked list has a cycle in it. This bug could be caught if you test your code with a linked list of size 2

Java:

public ListNode reverseList(ListNode head) {
    if (head == null || head.next == null) return head;
    ListNode p = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return p;
}
// https://leetcode.com/problems/reverse-linked-list/solution/

C++:

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (!head || !(head -> next)) {
            return head;
        }
        ListNode* node = reverseList(head -> next);
        head -> next -> next = head;
        head -> next = NULL;
        return node;
    }
};
// https://leetcode.com/problems/reverse-linked-list/discuss/58130/C%2B%2B-Iterative-and-Recursive

Python:

class Solution:
# @param {ListNode} head
# @return {ListNode}
def reverseList(self, head):
    return self._reverse(head)

def _reverse(self, node, prev=None):
    if not node:
        return prev
    n = node.next
    node.next = prev
    return self._reverse(n, node)
// https://leetcode.com/problems/reverse-linked-list/discuss/58127/Python-Iterative-and-Recursive-Solution