1472. Design Browser History
Description
You have a browser of one tab where you start on the homepage
and you can visit another url
, get back in the history number of steps
or move forward in the history number of steps
.
Implement the BrowserHistory
class:
BrowserHistory(string homepage)
Initializes the object with thehomepage
of the browser.void visit(string url)
Visitsurl
from the current page. It clears up all the forward history.string back(int steps)
Movesteps
back in history. If you can only returnx
steps in the history andsteps > x
, you will return onlyx
steps. Return the currenturl
after moving back in history at moststeps
.string forward(int steps)
Movesteps
forward in history. If you can only forwardx
steps in the history andsteps > x
, you will forward onlyx
steps. Return the currenturl
after forwarding in history at moststeps
.
Example:
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"] Explanation: BrowserHistory browserHistory = new BrowserHistory("leetcode.com"); browserHistory.visit("google.com"); // You are in "leetcode.com". Visit "google.com" browserHistory.visit("facebook.com"); // You are in "google.com". Visit "facebook.com" browserHistory.visit("youtube.com"); // You are in "facebook.com". Visit "youtube.com" browserHistory.back(1); // You are in "youtube.com", move back to "facebook.com" return "facebook.com" browserHistory.back(1); // You are in "facebook.com", move back to "google.com" return "google.com" browserHistory.forward(1); // You are in "google.com", move forward to "facebook.com" return "facebook.com" browserHistory.visit("linkedin.com"); // You are in "facebook.com". Visit "linkedin.com" browserHistory.forward(2); // You are in "linkedin.com", you cannot move forward any steps. browserHistory.back(2); // You are in "linkedin.com", move back two steps to "facebook.com" then to "google.com". return "google.com" browserHistory.back(7); // You are in "google.com", you can move back only one step to "leetcode.com". return "leetcode.com"
Constraints:
1 <= homepage.length <= 20
1 <= url.length <= 20
1 <= steps <= 100
homepage
andurl
consist of '.' or lower case English letters.- At most
5000
calls will be made tovisit
,back
, andforward
.
Solutions
Solution 1: Two Stacks
We can use two stacks, $\textit{stk1}$ and $\textit{stk2}$, to store the back and forward pages, respectively. Initially, $\textit{stk1}$ contains the $\textit{homepage}$, and $\textit{stk2}$ is empty.
When calling $\text{visit}(url)$, we add $\textit{url}$ to $\textit{stk1}$ and clear $\textit{stk2}$. The time complexity is $O(1)$.
When calling $\text{back}(steps)$, we pop the top element from $\textit{stk1}$ and push it to $\textit{stk2}$. We repeat this operation $steps$ times until the length of $\textit{stk1}$ is $1$ or $steps$ is $0$. Finally, we return the top element of $\textit{stk1}$. The time complexity is $O(\textit{steps})$.
When calling $\text{forward}(steps)$, we pop the top element from $\textit{stk2}$ and push it to $\textit{stk1}$. We repeat this operation $steps$ times until $\textit{stk2}$ is empty or $steps$ is $0$. Finally, we return the top element of $\textit{stk1}$. The time complexity is $O(\textit{steps})$.
The space complexity is $O(n)$, where $n$ is the length of the browsing history.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 |
|