**source:**https://leetcode.com/problems/reverse-linked-list**C/C++**

**Solution to LeetCode**problem

**206**.

**Reverse Linked List**.

## Problem

Given the `head`

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

## Examples

**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

- The number of nodes in the list is the range
`[0, 5000]`

. `-5000 <= Node.val <= 5000`

## Solution

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:

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/**
* 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:

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/**
* 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;
}
};