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.

class BrowserHistory {

    private int index; // 当前页面对应的 history 下标
    private final List<String> history; // 访问记录

    public BrowserHistory(String homepage) {
        index = 0;
        history = new ArrayList<>();
        history.add(homepage);
    }

    public void visit(String url) {
        // 若执行过 back, 那么 index 一定位于 history.size() - 1 之前
        // 此时可以进行 forward
        // 按照题意, 执行 visited 之后会抹除可以 forward 的页面
        // 所以把那些页面 remove 掉
        while (history.size() - 1 != index) {
            history.remove(history.size() - 1);
        }
        history.add(url);
        index = history.size() - 1;
    }

    public String back(int steps) {
        // 将 index 向前移动 steps 个位置即可
        steps = Math.min(steps, index);
        index -= steps;
        return history.get(index);
    }

    public String forward(int steps) {
        // 将 index 向后移动 steps 个位置即可
        steps = Math.min(steps, history.size() - 1 - index);
        index += steps;
        return history.get(index);
    }
}

关键点：执行visit之后如果current index不在头顶，那就要把curindex到头顶的先remove掉，再把url添加进去