栈的实现
225. Implement Stack using Queues
Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push
, top
, pop
, and empty
).
Implement the MyStack
class:
void push(int x)
Pushes element x to the top of the stack.int pop()
Removes the element on the top of the stack and returns it.int top()
Returns the element on the top of the stack.boolean empty()
Returnstrue
if the stack is empty,false
otherwise.
Notes:
- You must use only standard operations of a queue, which means that only
push to back
,peek/pop from front
,size
andis empty
operations are valid. - Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue's standard operations.
Example 1:
Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]
Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False
Constraints:
1 <= x <= 9
- At most
100
calls will be made topush
,pop
,top
, andempty
. - All the calls to
pop
andtop
are valid.
Follow-up: Can you implement the stack using only one queue?
思路
使用一个队列或两个队列都可以,使用一个队列时,弹出操作需要先弹出size-1
个元素再加入,再弹出所要弹出的元素。
C++解法
class MyStack {
public:
queue<int> q;
MyStack() {
}
void push(int x) {
q.push(x);
}
int pop() {
int size = q.size();
size--;
while(size--){
q.push(q.front());
q.pop();
}
int result = q.front();
q.pop();
return result;
}
int top() {
return q.back();
}
bool empty() {
return q.size() == 0;
}
};
/**
* 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();
*/