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| 1 | +#!/bin/python3 |
| 2 | + |
| 3 | +import math |
| 4 | +import os |
| 5 | +import random |
| 6 | +import re |
| 7 | +import sys |
| 8 | + |
| 9 | +class SinglyLinkedListNode: |
| 10 | + def __init__(self, node_data): |
| 11 | + self.data = node_data |
| 12 | + self.next = None |
| 13 | + |
| 14 | +class SinglyLinkedList: |
| 15 | + def __init__(self): |
| 16 | + self.head = None |
| 17 | + self.tail = None |
| 18 | + |
| 19 | + def insert_node(self, node_data): |
| 20 | + node = SinglyLinkedListNode(node_data) |
| 21 | + |
| 22 | + if not self.head: |
| 23 | + self.head = node |
| 24 | + else: |
| 25 | + self.tail.next = node |
| 26 | + |
| 27 | + |
| 28 | + self.tail = node |
| 29 | + |
| 30 | +def print_singly_linked_list(node, sep, fptr): |
| 31 | + while node: |
| 32 | + fptr.write(str(node.data)) |
| 33 | + |
| 34 | + node = node.next |
| 35 | + |
| 36 | + if node: |
| 37 | + fptr.write(sep) |
| 38 | + |
| 39 | +# Complete the findMergeNode function below. |
| 40 | + |
| 41 | +# |
| 42 | +# For your reference: |
| 43 | +# |
| 44 | +# SinglyLinkedListNode: |
| 45 | +# int data |
| 46 | +# SinglyLinkedListNode next |
| 47 | +# |
| 48 | +# |
| 49 | +def len_list(head1, head2): |
| 50 | + current1, count1 = head1, 0 |
| 51 | + current2, count2 = head2, 0 |
| 52 | + while current1: |
| 53 | + count1 += 1 |
| 54 | + print("current1:", current1.data, end=',') |
| 55 | + current1 = current1.next |
| 56 | + print() |
| 57 | + while current2: |
| 58 | + count2 += 1 |
| 59 | + print("current2:", current2.data, end=',') |
| 60 | + current2 = current2.next |
| 61 | + return (count1, count2) |
| 62 | + |
| 63 | +def findMergeNode(head1, head2): |
| 64 | + # O(n) time complexity to calculate the length of linked list |
| 65 | + # Also, traversing the linked list in above function for debugging |
| 66 | + count1, count2 = len_list(head1, head2) |
| 67 | + print("\n", count1, count2) |
| 68 | + # Assuming linked list by current1 is smaller in length compared to that by current2 |
| 69 | + # Thus, calculating diff by count2-count1 |
| 70 | + current1, current2 = head1, head2 |
| 71 | + d = count2 - count1 |
| 72 | + # if count1>count2, making diff positive, swapping pointers value to ensure the |
| 73 | + # largest linked list is always denoted by pointer current2. |
| 74 | + if count1 > count2: |
| 75 | + d = count1 - count2 |
| 76 | + current1, current2 = current2, current1 |
| 77 | + print("d:{}".format(d)) |
| 78 | + # O(j) time to reach the equidistant point from merge node |
| 79 | + for i in range(d): |
| 80 | + print("current2, before iter:{}".format(current2.data), end=',') |
| 81 | + current2 = current2.next |
| 82 | + print("\ncurrent2, after iter:{}".format(current2.data), end=',') |
| 83 | + print() |
| 84 | + # pointers are at equidistant node from the merge node, iterating until we get the |
| 85 | + # same node in both the pointers. |
| 86 | + while current1 and current2: |
| 87 | + print("current1:{}, current2:{}".format(current1.data, current2.data)) |
| 88 | + if current1 == current2: |
| 89 | + return current1.data |
| 90 | + current1 = current1.next |
| 91 | + current2 = current2.next |
| 92 | + """ |
| 93 | + Takes Time=O(n+m), Space=O(n) |
| 94 | + dic = {} |
| 95 | + current = head1 |
| 96 | + while current: |
| 97 | + dic[current] = 1 |
| 98 | + current = current.next |
| 99 | + current = head2 |
| 100 | + while current: |
| 101 | + if current in dic: |
| 102 | + return current.data |
| 103 | + current = current.next |
| 104 | + """ |
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