leetcode 232. Implement Queue using Stacks

题目描述：

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).

解题思路：

本题比较简单，主要考察对队列和栈的理解，然后注意摆脱只能使用一个栈的思维定势就可以了

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

// Removes the element from in front of queue.
void pop(void) {
if(out.empty())
trans();
out.pop();
}

// Get the front element.
int peek(void) {
if(out.empty())
trans();
return out.top();
}

// Return whether the queue is empty.
bool empty(void) {
return out.empty() && in.empty();
}
private:
stack<int> in, out;
void trans(){
while(!in.empty()){
out.push(in.top());
in.pop();
}
}
};

另附上leetcode上的题解，上面有对时空复杂度进行分析，时空复杂度在本题不是很直观(这涉及到均摊复杂度的概念 Amortized
Analysis )，可以看一下～