232. Implement Queue using Stacks

问题：

设计数据结构，使用stack实现queue。

实现以下功能：

• 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.

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.

　　

解法：双栈->实现队列

参考labuladong解析

重点在：

查队列头peek 和 出队pop

实现方法：

• 若s2中有元素，那么优先出队
• 若s2中无元素，那么s1.pop排空->s2.push

代码参考：

class MyQueue {
public:
    /** Initialize your data structure here. */
    MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    void push(int x) {
        s1.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        int res = peek();
        s2.pop();
        return res;
    }
    
    /** Get the front element. */
    int peek() {
        if (s2.empty()) {
            while(!s1.empty()) {
                s2.push(s1.top());
                s1.pop();
            }
        }
        return s2.top();
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return s2.empty() && s1.empty();
    }
private:
    stack<int> s1, 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();
 */