题目:
Implement the following operations of a stack using queues.
- push(x) -- Push element x onto stack.
- pop() -- Removes the element on top of the stack.
- top() -- Get the top element.
- empty() -- Return whether the stack is empty.
- You must use only standard operations of a queue -- which means only
push to back
,peek/pop from front
,size
, andis empty
operations are valid. - Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
- You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
思路:
我们首先看看队列和栈的区别:队列是先进先出,而栈是后进先出。可以看出算法实现的关键就在于在队列中存储元素的时候,将最近push进来的元素放在队列首位还是末尾了。根据这个区别,就可以有两种实现方法:
1、懒惰实现:顾名思义,就是新加入一个元素之后,我们懒惰第将它放在队列末尾。这就导致一个问题,当需要pop的时候,我们此时需要pop掉的是队列中的最后一个元素,而这个操作是队列这种数据结构所不支持的。所以在懒惰实现的pop中,我们需要首先把除队末元素之外的所有队列元素都移出队列,然后重新添加,这样就可以保证原来的队末元素此时位于队列头部了,可以直接pop。
2、勤奋实现:顾名思义,就是新push一个元素进来的时候,我们想办法把它置于队列头部,这样当需要pop的时候直接pop出来队列头部就可以了。所以这种“勤奋实现”的关键在于push函数的实现。事实上它和“懒惰实现”的pop函数完全一样:先把新元素加在队列元素的末尾,然后把所有除此之外的元素顺次移除,并添加到队列的末尾。
代码:
1、懒惰实现:
class MyStack {
public:
/** Initialize your data structure here. */
MyStack() {
}
/** Push element x onto stack. */
void push(int x) {
que.push(x);
}
/** Removes the element on top of the stack and returns that element. */
int pop() {
int size = que.size();
for (int i = 0; i < size - 1; ++i) {
que.push(que.front());
que.pop();
}
int ret = que.front();
que.pop();
return ret;
}
/** Get the top element. */
int top() {
return que.back();
}
/** Returns whether the stack is empty. */
bool empty() {
return que.empty();
}
private:
queue<int> que;
};
/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* bool param_4 = obj.empty();
*/
2、勤奋实现:
class MyStack {
public:
/** Initialize your data structure here. */
MyStack() {
}
/** Push element x onto stack. */
void push(int x) {
que.push(x);
for (int i = 0; i < que.size() - 1; ++i) {
que.push(que.front());
que.pop();
}
}
/** Removes the element on top of the stack and returns that element. */
int pop() {
int ret = que.front();
que.pop();
return ret;
}
/** Get the top element. */
int top() {
return que.front();
}
/** Returns whether the stack is empty. */
bool empty() {
return que.empty();
}
private:
queue<int> que;
};
/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* bool param_4 = obj.empty();
*/