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    LeetCode link: 206. Reverse Linked List, difficulty: Easy.

    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: [1,2]

    Output: [2,1]

    Example 3:

    Input: []

    Output: []

    Constraints:

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

    Intuition
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    1. To solve this problem, we only need to define two variables: current and previous. How do we inverse two node?

      Click to view the answer

      current.next = previous is the inversion.

    2. Which should be the loop condition? while (current != null) or while (current.next != null)?

      Click to view the answer

      It is while (current != null), because the operation to be performed is current.next = previous.

    Steps

    1. Traverse all nodes.

      previous = null
current = head

while (current != null) {
    current = current.next
}

    2. Add current.next = previous.

      previous = null
current = head

while (current != null) {
    tempNext = current.next
    current.next = previous
    current = tempNext
}

    3. Currently previous is always null, we need to change it: previous = current.

      previous = null
current = head

while (current != null) {
    tempNext = current.next
    current.next = previous
    previous = current
    current = tempNext
}

    Complexity

    Time complexity

    O(N)

    Space complexity

    O(1)

    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Java # 

    /**
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode previous = null;
        var current = head;

        while (current != null) {
            var tempNext = current.next;
            current.next = previous;
            previous = current;
            current = tempNext;
        }

        return previous;
    }
}

    Python # 

    # class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        previous = None
        current = head

        while current:
            temp_next = current.next
            current.next = previous
            previous = current
            current = temp_next

        return previous

    C++ # 

    /**
 * 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* previous = nullptr;
        ListNode* current = head;

        while (current) {
            auto temp_next = current->next;
            current->next = previous;
            previous = current;
            current = temp_next;
        }

        return previous;
    }
};

    JavaScript # 

    /**
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
var reverseList = function (head) {
  let previous = null
  let current = head

  while (current != null) {
    const tempNext = current.next
    current.next = previous
    previous = current
    current = tempNext
  }

  return previous
};

    C# # 

    /**
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution
{
    public ListNode ReverseList(ListNode head)
    {
        ListNode previous = null;
        ListNode current = head;

        while (current != null)
        {
            var tempNext = current.next;
            current.next = previous;
            previous = current;
            current = tempNext;
        }

        return previous;
    }
}

    Go # 

    /**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
    var previous *ListNode
    current := head

    for current != nil {
        tempNext := current.Next
        current.Next = previous
        previous = current
        current = tempNext
    }

    return previous
}

    Ruby # 

    # class ListNode
#     attr_accessor :val, :next
# 
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end

def reverse_list(head)
  previous = nil
  current = head

  while current
    temp_next = current.next
    current.next = previous
    previous = current
    current = temp_next
  end

  previous
end
    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Other languages

    Welcome to contribute code to our GitHub repository, thanks! The location of this solution is 206. Reverse Linked List.
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