LeetCode #206: Reverse Linked List (C/C++).

easy

source: https://leetcode.com/problems/reverse-linked-list
C/C++ Solution to LeetCode problem 206. Reverse Linked List.

Problem

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

Examples

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Example 1:

Input: [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

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

Solution

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We just need to point each node to the previous node instead of the next one, and the last node becomes the head.
Two solutions, one using a while loop and the other recursive.

Solution 1:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* reverseList(ListNode* head) {
    ListNode* p = nullptr;
    ListNode* c = head;

    while (c) {
      ListNode* n = c->next;
      c->next = p;
      p = c;
      c = n;
    }

    return p;
  }
};

Solution 2:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* reverseList(ListNode* head) {
    if (head == nullptr || head->next == nullptr)
      return head;

    ListNode* tmpHead = reverseList(head->next);
    head->next->next = head;
    head->next = nullptr;

    return tmpHead;
  }
};