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Copy pathreverse llist.py
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reverse llist.py
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# Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty.
# Example
# references the list
# Manipulate the pointers of each node in place and return , now referencing the head of the list .
# Function Description
# Complete the reverse function in the editor below.
# reverse has the following parameter:
# SinglyLinkedListNode pointer head: a reference to the head of a list
# Returns
# SinglyLinkedListNode pointer: a reference to the head of the reversed list
# Input Format
# The first line contains an integer , the number of test cases.
# Each test case has the following format:
# The first line contains an integer , the number of elements in the linked list.
# Each of the next lines contains an integer, the values of the elements in the linked list.
# Constraints
# , where is the element in the list.
# import math
# import os
# import random
# import re
# import sys
# class SinglyLinkedListNode:
# def __init__(self, node_data):
# self.data = node_data
# self.next = None
# class SinglyLinkedList:
# def __init__(self):
# self.head = None
# self.tail = None
# def insert_node(self, node_data):
# node = SinglyLinkedListNode(node_data)
# if not self.head:
# self.head = node
# else:
# self.tail.next = node
# self.tail = node
# def print_singly_linked_list(node, sep, fptr):
# while node:
# fptr.write(str(node.data))
# node = node.next
# if node:
# fptr.write(sep)
def reverse(llist):
# Write your code here
head = llist
if not head:
return
pre = None
cur = head
while cur:
next = cur.next
cur.next = pre
pre = cur
cur = next
head = pre
return head
# if __name__ == '__main__':
# fptr = open(os.environ['OUTPUT_PATH'], 'w')
# tests = int(input())
# for tests_itr in range(tests):
# llist_count = int(input())
# llist = SinglyLinkedList()
# for _ in range(llist_count):
# llist_item = int(input())
# llist.insert_node(llist_item)
# llist1 = reverse(llist.head)
# print_singly_linked_list(llist1, ' ', fptr)
# fptr.write('\n')
# fptr.close()