leetcode--设计浏览器历史记录
题目是LeetCode第192场周赛的第三题，链接：1472. 设计浏览器历史记录。具体描述为：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1:
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"]
很明显，这道题用栈来解决就很符合要求。这里需要用到两个栈，一个栈backHistory用来保存回退历史，另一个栈forwardHistory用来保存前进的历史。具体地说：
visit(url)时，把访问的url入backHistory，同时forwardHistory清空back(steps)时，如果steps大于backHistory.size()-1（我们需要保证homepage总是存在于backHistory中），那么steps重置为backHistory.size()-1，然后将backHistory中栈顶的steps个url都加入到forwardHistory中去，返回backHistory的栈顶元素forward(steps)时，如果steps大于forwardHistory.size()（因为forwardHistory可以为空），那么steps重置为forwardHistory.size()，然后将forwardHistory中栈顶的steps个url都加入到backHistory中去，返回forwardHistory的栈顶元素
JAVA版代码如下：
class BrowserHistory {
private Stack backHistory;
private Stack forwardHistory;
public BrowserHistory(String homepage) {
backHistory = new Stack<>();
backHistory.push(homepage);
forwardHistory = new Stack<>();
}
public void visit(String url) {
backHistory.push(url);
while (!forwardHistory.empty()) {
forwardHistory.pop();
}
}
public String back(int steps) {
steps = Math.min(steps, backHistory.size() - 1);
for (int i = 0; i < steps; ++i) {
forwardHistory.push(backHistory.pop());
}
return backHistory.peek();
}
public String forward(int steps) {
steps = Math.min(steps, forwardHistory.size());
for (int i = 0; i < steps; ++i) {
backHistory.push(forwardHistory.pop());
}
return backHistory.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);
*/
提交结果如下:
另一种方法是直接用一个链表保存url，同时用一个索引变量idx标识当前访问的位置，这么做除了在visit的时候需要remove掉idx之后的所有url之外，其他情况下都只需要改变idx，而不用进行remove。
JAVA版代码如下：
class BrowserHistory {
private int idx;
private List history;
public BrowserHistory(String homepage) {
idx = 0;
history = new ArrayList<>();
history.add(homepage);
}
public void visit(String url) {
int size = history.size();
int removeCount = size - idx - 1;
for (int i = 0; i < removeCount; ++i) {
history.remove(size - 1 - i);
}
history.add(url);
++idx;
}
public String back(int steps) {
idx = Math.max(0, idx - steps);
return history.get(idx);
}
public String forward(int steps) {
idx = Math.min(idx + steps, history.size() - 1);
return history.get(idx);
}
}
/**
* 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);
*/
提交结果如下:
Python版代码如下：
class BrowserHistory:
def __init__(self, homepage: str):
self.history = [homepage]
self.idx = 0
def visit(self, url: str) -> None:
count = len(self.history) - 1 - self.idx
for _ in range(count):
self.history.pop()
self.history.append(url)
self.idx += 1
def back(self, steps: int) -> str:
self.idx = max(0, self.idx - steps)
return self.history[self.idx]
def forward(self, steps: int) -> str:
self.idx = min(len(self.history) - 1, self.idx + steps)
return self.history[self.idx]
# 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)
提交结果如下:
