CODEWITHSUNDEEP
pythonlinked-list
Seungmin Lee
Seungmin Lee

Reputation: 51

dictionary question in python regarding leetcode#160

This is a leetcode question 160. intersection of two linked list, where Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. https://leetcode.com/problems/intersection-of-two-linked-lists/

if map[headB]==1 is giving a runtime error in Leetcode. I know if i use 'if headB in map' it works fine, but i want to know why 'if map[headB]==1' gives error. Thank you.

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

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
        
        map=dict()
        
        while (headA):
            map[headA]=1
            headA=headA.next
        
        while (headB):
            if map[headB]==1: #why this doesn't work? ('if headB in map' works fine)
                return headB
            headB=headB.next
        
        return None

Upvotes: 1

Views: 183

Answers (1)

Edmund
Edmund

Reputation: 36

Because if key headB doesn't exist in map, map[headB] will throw a KeyError. It cannot compare something that doesn't exist to one.

Also, you might want to use some other name than just "map", because it shadows the built-in Python function map().

Upvotes: 1

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