237. Delete Node in a Linked List

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Supposed the linked list is1 -> 2 -> 3 -> 4and you are given the third node with value3, the linked list should become1 -> 2 -> 4after calling your function.

Thoughts:

1. Copy the node value of the next node to the current node
2. Optional: free the next node

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        *node = *node->next;
    }
};

Code (freeing space)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        auto next = node -> next;
        *node = *next;
        delete next;
    }
};

Code (Java / C#)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void deleteNode(ListNode node) {
        node.val = node.next.val;
        node.next = node.next.next;
    }
}

Code (Python)

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        node.val = node.next.val;
        node.next= node.next.next;

Code (JavaScript)

var deleteNode = function(node) {
    node.val = node.next.val;
    node.next = node.next.next;
};

Code (Ruby)

# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val)
#         @val = val
#         @next = nil
#     end
# end

# @param {ListNode} node
# @return {Void} Do not return anything, modify node in-place instead.
def delete_node(node)
    node.val = node.next.val;
    node.next = node.next.next;
end

Special Thanks to stefanpochmann's solution for the reference