[leetcode] 232. Implement Queue using Stacks

Implement the following operations of a queue using stacks.
push(x) -- Push element x to the back of queue.
pop() -- Removes the element from in front of queue.
peek() -- Get the front element.
empty() -- Return whether the queue is empty.
Notes:
You must use only standard operations of a stack -- which means only 

push
to top

, 

peek/pop from top

, 

size

,
and 

is empty

 operations are valid.
Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
解法一：

简单的说就是改变push函数。讲一个要存成1,2,3,4的queue，变成1,2,3,4的stack。于是，我们有q.pop() <-> s.pop(), q.peek() <->s.top(), q.empty()<->s.empty(). 在push()中，需要一个辅助stack去存储后的stack。

class Queue {
public:
// Push element x to the back of queue.
void push(int x) {
stack<int> tmp;
while(!s.empty()){
tmp.push(s.top());
s.pop();
}
tmp.push(x);
while(!tmp.empty()){
s.push(tmp.top());
tmp.pop();
}
}

// Removes the element from in front of queue.
void pop(void) {
s.pop();
}

// Get the front element.
int peek(void) {
return s.top();
}

// Return whether the queue is empty.
bool empty(void) {
return s.empty();
}

private:
stack<int> s;
};

解法二：

思路是维护两个stack, _new和_old。新进来的数都push到_new中，需要pop()或者peek()的时候，如果_old为空，将_new的数倒序放到_old里面，然后对_old里最top的数操作。如果_old不为空，直接在其top上操作。

class Queue {
public:
// Push element x to the back of queue.
void push(int x) {
_new.push(x);

}

// Removes the element from in front of queue.
void pop(void) {
shiftStack();
_old.pop();
}

// Get the front element.
int peek(void) {
shiftStack();
return _old.top();
}

// Return whether the queue is empty.
bool empty(void) {
return _new.empty() && _old.empty();

}

void shiftStack(){
if(_old.empty()){
while(!_new.empty()){
_old.push(_new.top());
_new.pop();
}

}
}

private:
stack<int> _new, _old;
};