[LeetCode] Implement Queue using Stacks

Implement Queue using Stacks

Implement the following operations of a queue using stacks.

• push(x) -- Push element x to the back of queue.
• pop() -- Removes the element from in front of queue.
• peek() -- Get the front element.
• empty() -- Return whether the queue is empty.

  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, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
  • You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

    解題思路：

    用兩個棧來模擬隊列。其中一個棧做緩存之用，另外一個棧只接受。push操作，直接放入接收之棧中。pop操作，若緩存棧存在數據，則從緩存棧中彈出。若緩存棧為空，則將接收棧的數據彈出並插入到緩存棧中，再從緩存棧中彈出。peek操作與pop操作類似。當且僅當兩個棧都為空時，empty才返回true。

     

    class Queue {
    public:
        // Push element x to the back of queue.
        void push(int x) {
            stacks[0].push(x);
        }
    
        // Removes the element from in front of queue.
        void pop(void) {
            if(stacks[1].empty()){
                while(!stacks[0].empty()){
                    stacks[1].push(stacks[0].top());
                    stacks[0].pop();
                }
            }
            stacks[1].pop();
        }
    
        // Get the front element.
        int peek(void) {
            if(stacks[1].empty()){
                while(!stacks[0].empty()){
                    stacks[1].push(stacks[0].top());
                    stacks[0].pop();
                }
            }
            return stacks[1].top();
        }
    
        // Return whether the queue is empty.
        bool empty(void) {
            return stacks[0].empty() && stacks[1].empty();
        }
        
    private:
        stack stacks[2];
    };

    
    
    

     

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