delete-node-in-a-linked-list.py

# -*- coding: utf-8 -*-
"""
Created on Sat Apr 25 02:42:13 2020

@author: johnoyegbite
"""
# SOLVED!
"""
Problem:
    Write a function to delete a node (except the tail) in a singly linked
    list, given only access to that node.

Example 1:
    Input: head = [4,5,1,9], node = 5
    Output: [4,1,9]
    Explanation:
        You are given the second node with value 5, the linked list should
        become 4 -> 1 -> 9 after calling your function.

Example 2:
    Input: head = [4,5,1,9], node = 1
    Output: [4,5,9]
    Explanation:
        You are given the third node with value 1, the linked list should
        become 4 -> 5 -> 9 after calling your function.

Note:
    The linked list will have at least two elements.
    All of the nodes' values will be unique.
    The given node will not be the tail and it will always be a valid node
    of the linked list.
    Do not return anything from your function.
"""


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


class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        # 1. copy the value of the next node to the current node
        # 2. point the current node to the upper node (which automatically
        #    delete/neglect the next node.)
        node.val = node.next.val
        node.next = node.next.next