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1472. 设计浏览器历史记录

English Version

题目描述

你有一个只支持单个标签页的 浏览器 ，最开始你浏览的网页是 homepage ，你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。

请你实现 BrowserHistory 类：

• BrowserHistory(string homepage) ，用 homepage 初始化浏览器类。
• void visit(string url) 从当前页跳转访问 url 对应的页面  。执行此操作会把浏览历史前进的记录全部删除。
• string back(int steps) 在浏览历史中后退 steps 步。如果你只能在浏览历史中后退至多 x 步且 steps > x ，那么你只后退 x 步。请返回后退 至多 steps 步以后的 url 。
• string forward(int steps) 在浏览历史中前进 steps 步。如果你只能在浏览历史中前进至多 x 步且 steps > x ，那么你只前进 x 步。请返回前进 至多 steps步以后的 url 。

 

示例：

输入：
["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]]
输出：
[null,null,null,null,"facebook.com","google.com","facebook.com",null,"linkedin.com","google.com","leetcode.com"]

解释：
BrowserHistory browserHistory = new BrowserHistory("leetcode.com");
browserHistory.visit("google.com");       // 你原本在浏览 "leetcode.com" 。访问 "google.com"
browserHistory.visit("facebook.com");     // 你原本在浏览 "google.com" 。访问 "facebook.com"
browserHistory.visit("youtube.com");      // 你原本在浏览 "facebook.com" 。访问 "youtube.com"
browserHistory.back(1);                   // 你原本在浏览 "youtube.com" ，后退到 "facebook.com" 并返回 "facebook.com"
browserHistory.back(1);                   // 你原本在浏览 "facebook.com" ，后退到 "google.com" 并返回 "google.com"
browserHistory.forward(1);                // 你原本在浏览 "google.com" ，前进到 "facebook.com" 并返回 "facebook.com"
browserHistory.visit("linkedin.com");     // 你原本在浏览 "facebook.com" 。 访问 "linkedin.com"
browserHistory.forward(2);                // 你原本在浏览 "linkedin.com" ，你无法前进任何步数。
browserHistory.back(2);                   // 你原本在浏览 "linkedin.com" ，后退两步依次先到 "facebook.com" ，然后到 "google.com" ，并返回 "google.com"
browserHistory.back(7);                   // 你原本在浏览 "google.com"， 你只能后退一步到 "leetcode.com" ，并返回 "leetcode.com"

 

提示：

• 1 <= homepage.length <= 20
• 1 <= url.length <= 20
• 1 <= steps <= 100
• homepage 和 url 都只包含 '.' 或者小写英文字母。
• 最多调用 5000 次 visit， back 和 forward 函数。

解法

方法一：栈

使用两个栈模拟前进与后退操作。

Python3

class BrowserHistory:

    def __init__(self, homepage: str):
        self.stk1 = []
        self.stk2 = []
        self.visit(homepage)

    def visit(self, url: str) -> None:
        self.stk1.append(url)
        self.stk2.clear()

    def back(self, steps: int) -> str:
        while steps and len(self.stk1) > 1:
            self.stk2.append(self.stk1.pop())
            steps -= 1
        return self.stk1[-1]

    def forward(self, steps: int) -> str:
        while steps and self.stk2:
            self.stk1.append(self.stk2.pop())
            steps -= 1
        return self.stk1[-1]


# 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)

Java

class BrowserHistory {
    private Deque<String> stk1 = new ArrayDeque<>();
    private Deque<String> stk2 = new ArrayDeque<>();

    public BrowserHistory(String homepage) {
        visit(homepage);
    }

    public void visit(String url) {
        stk1.push(url);
        stk2.clear();
    }

    public String back(int steps) {
        for (; steps > 0 && stk1.size() > 1; --steps) {
            stk2.push(stk1.pop());
        }
        return stk1.peek();
    }

    public String forward(int steps) {
        for (; steps > 0 && !stk2.isEmpty(); --steps) {
            stk1.push(stk2.pop());
        }
        return stk1.peek();
    }
}

/**
 * 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);
 */

C++

class BrowserHistory {
public:
    stack<string> stk1;
    stack<string> stk2;

    BrowserHistory(string homepage) {
        visit(homepage);
    }

    void visit(string url) {
        stk1.push(url);
        stk2 = stack<string>();
    }

    string back(int steps) {
        for (; steps && stk1.size() > 1; --steps) {
            stk2.push(stk1.top());
            stk1.pop();
        }
        return stk1.top();
    }

    string forward(int steps) {
        for (; steps && !stk2.empty(); --steps) {
            stk1.push(stk2.top());
            stk2.pop();
        }
        return stk1.top();
    }
};

/**
 * 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);
 */

Go

type BrowserHistory struct {
	stk1 []string
	stk2 []string
}

func Constructor(homepage string) BrowserHistory {
	t := BrowserHistory{[]string{}, []string{}}
	t.Visit(homepage)
	return t
}

func (this *BrowserHistory) Visit(url string) {
	this.stk1 = append(this.stk1, url)
	this.stk2 = []string{}
}

func (this *BrowserHistory) Back(steps int) string {
	for i := 0; i < steps && len(this.stk1) > 1; i++ {
		this.stk2 = append(this.stk2, this.stk1[len(this.stk1)-1])
		this.stk1 = this.stk1[:len(this.stk1)-1]
	}
	return this.stk1[len(this.stk1)-1]
}

func (this *BrowserHistory) Forward(steps int) string {
	for i := 0; i < steps && len(this.stk2) > 0; i++ {
		this.stk1 = append(this.stk1, this.stk2[len(this.stk2)-1])
		this.stk2 = this.stk2[:len(this.stk2)-1]
	}
	return this.stk1[len(this.stk1)-1]
}

/**
 * Your BrowserHistory object will be instantiated and called as such:
 * obj := Constructor(homepage);
 * obj.Visit(url);
 * param_2 := obj.Back(steps);
 * param_3 := obj.Forward(steps);
 */

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