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?
Companies: Google, Amazon, Bloomberg
Related Topics:
Stack, Design, Queue
Similar Questions:
// OJ: https://leetcode.com/problems/implement-stack-using-queues
// Author: github.com/lzl124631x
// Time: O(N) for push, O(1) for others
// Space: O(1)
class MyStack {
queue<int> q;
public:
MyStack() {}
void push(int x) {
q.push(x);
for (int cnt = q.size() - 1; cnt > 0; --cnt) {
q.push(q.front());
q.pop();
}
}
int pop() {
int ans = q.front();
q.pop();
return ans;
}
int top() {
return q.front();
}
bool empty() {
return q.empty();
}
};
// OJ: https://leetcode.com/problems/implement-stack-using-queues/
// Author: github.com/lzl124631x
// Time: O(N) for pop, O(1) for others
// Space: O(1)
class MyStack {
queue<int> q;
int topVal;
public:
MyStack() {}
void push(int x) {
q.push(x);
topVal = x;
}
int pop() {
int ans = topVal;
for (int i = 0, n = q.size(); i < n - 1; ++i) {
topVal = q.front();
q.push(topVal);
q.pop();
}
q.pop();
return ans;
}
int top() {
return topVal;
}
bool empty() {
return q.empty();
}
};