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# 1472. Design Browser History

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

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`.

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

模拟浏览器的访问，前进和后退操作，back(int step)把浏览记录后退step步，forward(int step)是前进step步，visit(url)会浏览url并清除所有它前面的网页。根据题意相当于在array上维持一个指针p不断跳，当调用visit时把p后插入新的url并把之后的清空。为了实现O(1)清空效果可以维持指针e指向当前上界。

```python
class BrowserHistory:

    def __init__(self, homepage: str):
        self.l = [homepage]
        self.p, self.e = 0, 0

    def visit(self, url: str) -> None:
        self.p += 1
        if self.p == len(self.l):
            self.l.append(url)
        else:
            self.l[self.p] = url
        self.e = self.p
        
    def back(self, steps: int) -> str:
        self.p = max(self.p - steps, 0)
        return self.l[self.p]

    def forward(self, steps: int) -> str:
        self.p = min(self.p + steps, self.e)
        return self.l[self.p]


# Your BrowserHistory object will be instantiated and called as such:
# obj = BrowserHistory(homepage)
# obj.visit(url)
# param_2 = obj.back(steps)
# param_3 = obj.forward(steps)
```