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?
class MyStack {
public:
queue<int>queue;
MyStack() {
}
void push(int x) {
queue.push(x);
}
int pop() {
int size=queue.size();
size--;
while(size--){
queue.push(queue.front());
queue.pop();
}
int result=queue.front();
queue.pop();
return result;
}
int top() {
return queue.back();
}
bool empty() {
return queue.empty();
}
};
注意:
1.两个队列也是可以的,但是一个queue显得更简洁,而且也很好理解
2.pop的时候,记住size要先-1
3.top是指队列最后进去的那个元素,使用的是back,不是front
标签:Queues,pop,stack,queue,push,using,225,top,empty From: https://blog.csdn.net/2301_80161204/article/details/140710910