Linked List - leetcode 예시로 이해

https://leetcode.com/problems/design-browser-history/

Design Browser History - LeetCode

class ListNOde(object):
	def __init__(self, val=0, next=None, prev=None):
    		self.val = val
			self.next = next
        	self.prev = prev
class BrowserHistory(object):
	def __init__(self, homepage):
    		self.head = self.current = ListNode(val=homepage)
	def visit(self,url):
    		self.current.next = ListNode(val=url, prev=self.current)
        	self.current = self.current.next
        	return None
    def back(self, steps):
    		while steps > 0 and self.current.prev != None:
        		steps -= 1
            	self.current = self.current.prev
        	return self.current.val
    def forward(self, steps):
    		while steps > 0 and self.current.next != None:
        		steps -= 1
            	self.current = self.current.next
        	return self.current.val

INPUT

["BrowserHistory","visit","visit","visit","back","back","forward","visit","forward","back","back"]
[["leetcode.com"],["google.com"],["facebook.com"],["youtube.com"],[1],[1],[1],["linkedin.com"],[2],[2],[7]]

OUTPUT

[null,null,null,null,"facebook.com","google.com","facebook.com",null,"linkedin.com","google.com","leetcode.com"]

클래식한 linked list의 좋은 문제였다.

내부동작

1.BrowserHistory("leetcode.com") 

homepage: leetcode.com

current : leetcode.com

head: leetcode.com

2.visit("goole.com")

val : google.com

prev : current(즉, leetcode.com)

current.next: google.com

3.back

받은 steps을 0이 되거나 prev가 None될 때(맨앞으로 와서 더이상 앞이 없을때)까지 while문 self.current 를 current.prev(한칸 앞으로 간다.)

4.forward

back과 과정은 같지만 앞으로간다.