Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).
Implement the MyQueue class:
void push(int x) Pushes element x to the back of the queue.
int pop() Removes the element from the front of the queue and returns it.
int peek() Returns the element at the front of the queue.
boolean empty() Returns true if the queue is empty, false otherwise.
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, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.
Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.
Explanation MyQueue myQueue = new MyQueue(); myQueue.push(1); // queue is: [1] myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue) myQueue.peek(); // return 1 myQueue.pop(); // return 1, queue is [2] myQueue.empty(); // return false
Constraints:
1 <= x <= 9
At most 100 calls will be made to push, pop, peek, and empty.
All the calls to pop and peek are valid.
Source code
Version 1
Idea: The frist method uses push() function to implement Queue. Use two Stacks, if s1 is emtpy, push x directly. If s1 is not empty, move all of elements in s1 to s2, and push x into the emtpy s1, finally, move all of elements in s2 back to s1.
classMyQueue { public: /** Initialize your data structure here. */ MyQueue() { } /** Push element x to the back of queue. */ voidpush(int x){ if(s1.empty()){ s1.push(x); }else{ while(!s1.empty()){ s2.push(s1.top()); s1.pop(); } s1.push(x); while(!s2.empty()){ s1.push(s2.top()); s2.pop(); } } } /** Removes the element from in front of queue and returns that element. */ intpop(){ int temp = s1.top(); s1.pop(); return temp; } /** Get the front element. */ intpeek(){ return s1.top(); } /** Returns whether the queue is empty. */ boolempty(){ return s1.empty(); } private: stack<int> s1; stack<int> s2; };
/** * Your MyQueue object will be instantiated and called as such: * MyQueue* obj = new MyQueue(); * obj->push(x); * int param_2 = obj->pop(); * int param_3 = obj->peek(); * bool param_4 = obj->empty(); */
Version 2
Idea: The second method uses pop() function to implement Queue. In pop() and peek(), always returns the element from s2. So, In the functions, it need to check whether s2 is empty or not. If s2 is empty, move all of elements in s1 to s2.
classMyQueue { public: /** Initialize your data structure here. */ MyQueue() { } /** Push element x to the back of queue. */ voidpush(int x){ s1.push(x); } /** Removes the element from in front of queue and returns that element. */ intpop(){ if(s2.empty()) moved(); int temp = s2.top(); s2.pop(); return temp; } /** Get the front element. */ intpeek(){ if(s2.empty()) moved(); return s2.top(); } /** Returns whether the queue is empty. */ boolempty(){ return s1.empty() && s2.empty(); } private: stack<int> s1; stack<int> s2; voidmoved(){ while(!s1.empty()){ s2.push(s1.top()); s1.pop(); } } };
/** * Your MyQueue object will be instantiated and called as such: * MyQueue* obj = new MyQueue(); * obj->push(x); * int param_2 = obj->pop(); * int param_3 = obj->peek(); * bool param_4 = obj->empty(); */