1472. 設計浏覽器曆史記錄](https://leetcode-cn.com/problems/design-browser-history/)
難度中等45收藏分享切換為英文接收動态回報
你有一個隻支援單個标簽頁的 浏覽器 ,最開始你浏覽的網頁是
homepage
,你可以通路其他的網站
url
,也可以在浏覽曆史中後退
steps
步或前進
steps
步。
請你實作
BrowserHistory
類:
-
,用BrowserHistory(string homepage)
初始化浏覽器類。homepage
-
從目前頁跳轉通路void visit(string url)
對應的頁面 。執行此操作會把浏覽曆史前進的記錄全部删除。url
-
在浏覽曆史中後退string back(int steps)
步。如果你隻能在浏覽曆史中後退至多steps
步且x
,那麼你隻後退steps > x
步。請傳回後退至多x
步以後的steps
。url
-
在浏覽曆史中前進string forward(int steps)
步。如果你隻能在浏覽曆史中前進至多steps
步且x
,那麼你隻前進steps > x
步。請傳回前進至多x
步以後的steps
。url
輸入:
["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]]
輸出:
[null,null,null,null,"facebook.com","google.com","facebook.com",null,"linkedin.com","google.com","leetcode.com"]
解釋:
BrowserHistory browserHistory = new BrowserHistory("leetcode.com");
browserHistory.visit("google.com"); // 你原本在浏覽 "leetcode.com" 。通路 "google.com"
browserHistory.visit("facebook.com"); // 你原本在浏覽 "google.com" 。通路 "facebook.com"
browserHistory.visit("youtube.com"); // 你原本在浏覽 "facebook.com" 。通路 "youtube.com"
browserHistory.back(1); // 你原本在浏覽 "youtube.com" ,後退到 "facebook.com" 并傳回 "facebook.com"
browserHistory.back(1); // 你原本在浏覽 "facebook.com" ,後退到 "google.com" 并傳回 "google.com"
browserHistory.forward(1); // 你原本在浏覽 "google.com" ,前進到 "facebook.com" 并傳回 "facebook.com"
browserHistory.visit("linkedin.com"); // 你原本在浏覽 "facebook.com" 。 通路 "linkedin.com"
browserHistory.forward(2); // 你原本在浏覽 "linkedin.com" ,你無法前進任何步數。
browserHistory.back(2); // 你原本在浏覽 "linkedin.com" ,後退兩步依次先到 "facebook.com" ,然後到 "google.com" ,并傳回 "google.com"
browserHistory.back(7); // 你原本在浏覽 "google.com", 你隻能後退一步到 "leetcode.com" ,并傳回 "leetcode.com"
class Node(object):
def __init__(self, pagename):
self.val = pagename
self.prev = None
self.next = None
class BrowserHistory:
def __init__(self, homepage: str):
self.head = Node(homepage)
def visit(self, url: str) -> None:
cur = self.head
node = Node(url)
node.prev = cur
cur.next = node
self.head = node
def back(self, steps: int) -> str:
while steps > 0 and self.head.prev:
self.head = self.head.prev
steps -= 1
return self.head.val
def forward(self, steps: int) -> str:
while steps > 0 and self.head.next:
steps -= 1
self.head = self.head.next
return self.head.val