LeetCode //C - 206. Reverse Linked List

206. Reverse Linked List

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

Example 1:

Input head = [1,2,3,4,5]
Output [5,4,3,2,1]

Example 2:

Input head = [1,2]
Output [2,1]

Example 3:

Input head = []
Output []

Constraints:

• The number of nodes in the list is the range [0, 5000].
• -5000 <= Node.val <= 5000

From: LeetCode
Link: 206. Reverse Linked List

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Solution:

Ideas：

• Initialize three pointers: prev (initially NULL), curr (pointing to the head of the list), and next (initially NULL).
• Iterate through the list. In each iteration:
  • Store the next node in next.
  • Reverse the current node’s next pointer to point to prev.
  • Move prev and curr one step forward.
• After the loop, prev will point to the new head of the reversed list.
• Update head to prev and return it.

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode *prev = NULL;
    struct ListNode *curr = head;
    struct ListNode *next = NULL;

    while (curr != NULL) {
        next = curr->next;  // Store next node
        curr->next = prev;  // Reverse current node's pointer
        prev = curr;        // Move pointers one position ahead
        curr = next;
    }

    head = prev;  // Update head to new first node
    return head;
}