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.
Notes:
You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is 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).
Update (2015-06-11):
The class name of the Java function had been updated to MyStack instead of Stack.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and all test cases.
Hide Tags Data Structure
分析:
由於棧是先進後出,隊列是先進先出,兩者出棧順序是相反的,因此,因此兩個隊列來實現一個棧,一個作爲中轉隊列,一個是目標隊列,二者不斷交換角色。過程描述如下:
在用隊列實現棧的出棧操作時,需要刪除隊隊尾的元素,那麼得將目標隊列的非隊尾元素出隊列存放到中轉隊列中,然後出隊原隊尾元素。此時目標隊列和中轉隊列角色互換。
對於壓棧操作,直接將元素放入目標隊列(非空)隊尾,若兩個隊列都爲空,那麼隨便選擇其中一個作爲目標的隊列即可。
對於判空,如過兩個隊列都爲空,那麼返回true,否則返回false。
對於top操作,直接返回非空隊列的隊尾元素即可。
思路還是挺簡單的,以下是C++實現代碼:
/**///////////////////////////0ms*/
class Stack {
public:
queue<int> q1,q2;
// Push element x onto stack.
void push(int x) {
if(q1.empty() && q2.empty() || !q1.empty())
q1.push(x);
else
q2.push(x);
}
// Removes the element on top of the stack.
void pop() {
int top = 0;
if(!Stack::empty()){
if(!q1.empty()){
int n = q1.size()-1;
while(n--)
{
q2.push(q1.front());
q1.pop();
}
q1.pop();
}
else{
int n = q2.size()-1;
while(n--)
{
q2.push(q2.front());
q2.pop();
}
q2.pop();
}
}
}
// Get the top element.
int top(){
if(!Stack::empty()){
if(!q1.empty())
return q1.back();
else
return q2.back();
}
}
// Return whether the stack is empty.
bool empty() {
if(q1.empty() && q2.empty())
return true;
return false;
}
};