232. Implement Queue using Stacks -E

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.

 

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

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.

 

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.

A:
  -----use two stack to simulate ----------

class MyQueue {
public:
    MyQueue() {
    }
    
    void push(int x) {
        inS.push(x);
    }
    
    int pop() {
        moveIn2Out();
        int res = outS.top();
        outS.pop();
        return res;
    }
    
    int peek() {
        moveIn2Out();
        return outS.top();
    }
    
    bool empty() {
        moveIn2Out();
        return outS.empty();
    }
private:
    stack<int> inS, outS;
     void moveIn2Out(){
        if(outS.empty()){
            while( ! inS.empty()){
                int tmp = inS.top();
                outS.push(tmp);
                inS.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();
 */
/**
 * 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();
 */
Errors:

    1:  用Stack的时候，用了vector的方法。很SB

    2：  只有当outS为空的时候，才能把inS都倒过去。否则是不对的