Question

Formatted question description: https://leetcode.ca/all/1472.html

 1472. 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.

Algorithm

In this question, we can use two Stacks to save the forward history list and the backward history list respectively, and use a variable to save the current url.

When executing the visit method, first clear the forward list, then save the current url to the back list, and then set the visited url as the current url.

When executing the back method, we first store the current url into the forward list, and then take an element from the backward list as the current url. Repeat this process until step back is completed or the back list is empty, the current url is the return value.

Similar to back, when executing the forward method, we first store the current url in the back list, and then take an element from the forward list as the current url. Repeat this process until the step forward is completed or the forward list is empty, and the current url is the return value.

Code

Java

public class Design_Browser_History {

    class BrowserHistory {

        Stack<String> back = new Stack<>();
        Stack<String> forward = new Stack<>();
        String current;

        public BrowserHistory(String homepage) {
            current = homepage;
        }

        public void visit(String url) {
            forward.clear();
            back.push(current);
            current = url;
        }

        public String back(int steps) {
            while (steps > 0 && back.size() > 0) {
                forward.push(current);
                current = back.pop();
                steps--;
            }
            return current;
        }

        public String forward(int steps) {
            while (steps > 0 && forward.size() > 0) {
                back.push(current);
                current = forward.pop();
                steps--;
            }
            return current;
        }
    }

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