206. Reverse Linked List

O(N) Time and O(1) space, iterative

Given the head of a singly linked list, reverse the list, and return the reversed list.

• Initialize three pointers prev as NULL, curr as head and next as NULL.
• Iterate through the linked list. In loop, do following.
  • Before changing next of current, store next node next = curr->next
  • Now change next of current This is where actual reversing happens curr->next = prev
  • Move prev and curr one step forward prev = curr curr = next

Code

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *prev = NULL,*next=NULL,*curr=head;
        while(curr!=NULL){
            next=curr->next;
            curr->next=prev;
            prev=curr;
            curr=next;
        }
        return prev;
    }
};

O(N) Time and O(1) space, recursive

1. Divide the list in two parts - first node and rest of the linked list.
2. Call reverse for the rest of the linked list.
3. Link rest to first.
4. Fix head pointer

Code

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head==NULL) return NULL;
        if(head->next==NULL) return head; // Make last node head
        
        ListNode* newHead = reverseList(head->next);
        
        head->next->next = head; // Actual reversal happens here
        head->next = NULL;
        
        return newHead;
    }
};

Best Video To visualize reverse linked list recursively