1472. 设计浏览器历史记录

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题目

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.

题目大意

设计一个只支持单个标签页的 浏览器 ，最开始浏览的网页是 homepage ，可以访问其他的网站 url ，也可以在浏览历史中后退 steps 步或前进 steps 步。

解题思路

使用栈来存储浏览器的访问历史，使用 cur_index 变量来存储当前访问的网址在栈中的位置。

• visit：
  • 先清空 cur_index 之后的历史记录；
  • 将 url 放入栈顶；
  • 将 cur_index 指向栈顶；
• back：比较栈中存储的历史记录数 n 和 steps 的大小，最多只能倒退 n 步
• forward: 比较 cur_index 之后的历史记录数 m 和 steps 的大小，最多只能前进 m 步

代码

class BrowserHistory {
	// @param {string} homepage
	constructor(homepage) {
		this.history = [homepage];
		this.cur_index = 0;
	}

	// @param {string} url
	// @return {void}
	visit(url) {
		// clear forward history
		this.history = this.history.slice(0, this.cur_index + 1);
		this.history.push(url);
		this.cur_index++;
	}

	// @param {number} steps
	// @return {string}
	back(steps) {
		this.cur_index = Math.max(0, this.cur_index - steps);
		return this.history[this.cur_index];
	}

	// @param {number} steps
	// @return {string}
	forward(steps) {
		this.cur_index = Math.min(this.history.length - 1, this.cur_index + steps);
		return this.history[this.cur_index];
	}
}
/**
 * Your BrowserHistory object will be instantiated and called as such:
 * var obj = new BrowserHistory(homepage)
 * obj.visit(url)
 * var param_2 = obj.back(steps)
 * var param_3 = obj.forward(steps)
 */

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