1472.Design Browser History
===

###### tags: `Medium`,`Array`,`Linked List`,`Stack`

[1472. Design Browser History](https://leetcode.com/problems/design-browser-history/)

### 題目描述

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 the `homepage` of the browser.
* `void visit(string url)` Visits `url` from the current page. It clears up all the forward history.
* `string back(int steps)` Move `steps` back in history. If you can only return `x` steps in the history and `steps > x`, you will return only `x` steps. Return the current `url` after moving back in history **at most** `steps`.
* `string forward(int steps)` Move `steps` forward in history. If you can only forward `x` steps in the history and `steps > x`, you will forward only `x` steps. Return the current `url` after forwarding in history **at most** `steps`.

### 範例

**Example 1:**

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` and `url` consist of '.' or lower case English letters.
* At most `5000` calls will be made to `visit`, `back`, and `forward`.

### 解答
#### Python

class BrowserHistory:

    def __init__(self, homepage: str):
        self.history = [homepage]
        self.cur = 0

    def visit(self, url: str) -> None:
        self.cur += 1
        self.history = self.history[:self.cur] + [url]

    def back(self, steps: int) -> str:
        self.cur = max(0, self.cur - steps)
        return self.history[self.cur]

    def forward(self, steps: int) -> str:
        self.cur = min(len(self.history) - 1, self.cur + steps)
        return self.history[self.cur]

> [name=Yen-Chi Chen][time=Sat, Mar 18, 2023]

### Reference

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