README_EN.md

1472. Design Browser History

中文文档

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 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.

Solutions

Python3

Using list.

class BrowserHistory:

    def __init__(self, homepage: str):
        self.urls = []
        self.cur = -1
        self.tail = -1
        self.visit(homepage)

    def visit(self, url: str) -> None:
        self.cur += 1
        if self.cur < len(self.urls):
            self.urls[self.cur] = url
        else:
            self.urls.append(url)
        self.tail = self.cur

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

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

# 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)

Using stacks.

class BrowserHistory:

    def __init__(self, homepage: str):
        self.s1 = []
        self.s2 = []
        self.cur = homepage

    def visit(self, url: str) -> None:
        self.s2.clear()
        self.s1.append(self.cur)
        self.cur = url

    def back(self, steps: int) -> str:
        while steps > 0 and self.s1:
            self.s2.append(self.cur)
            self.cur = self.s1.pop()
            steps -= 1
        return self.cur

    def forward(self, steps: int) -> str:
        while steps > 0 and self.s2:
            self.s1.append(self.cur)
            self.cur = self.s2.pop()
            steps -= 1
        return self.cur


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

Java

Using list.

class BrowserHistory {
    private List<String> urls;
    private int cur = -1;
    private int tail = -1;

    public BrowserHistory(String homepage) {
        urls = new ArrayList<>();
        visit(homepage);
    }

    public void visit(String url) {
        ++cur;
        if (cur < urls.size()) {
            urls.set(cur, url);
        } else {
            urls.add(url);
        }
        tail = cur;
    }

    public String back(int steps) {
        cur = Math.max(0, cur - steps);
        return urls.get(cur);
    }

    public String forward(int steps) {
        cur = Math.min(tail, cur + steps);
        return urls.get(cur);
    }
}

/**
 * Your BrowserHistory object will be instantiated and called as such:
 * BrowserHistory obj = new BrowserHistory(homepage);
 * obj.visit(url);
 * String param_2 = obj.back(steps);
 * String param_3 = obj.forward(steps);
 */

Using stacks.

class BrowserHistory {
    private Deque<String> s1;
    private Deque<String> s2;
    private String cur;

    public BrowserHistory(String homepage) {
        s1 = new ArrayDeque<>();
        s2 = new ArrayDeque<>();
        cur = homepage;
    }

    public void visit(String url) {
        s2.clear();
        s1.push(cur);
        cur = url;
    }

    public String back(int steps) {
        while (steps > 0 && !s1.isEmpty()) {
            s2.push(cur);
            cur = s1.pop();
            --steps;
        }
        return cur;
    }

    public String forward(int steps) {
        while (steps > 0 && !s2.isEmpty()) {
            s1.push(cur);
            cur = s2.pop();
            --steps;
        }
        return cur;
    }
}

/**
 * Your BrowserHistory object will be instantiated and called as such:
 * BrowserHistory obj = new BrowserHistory(homepage);
 * obj.visit(url);
 * String param_2 = obj.back(steps);
 * String param_3 = obj.forward(steps);
 */

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