【LeetCode】232. Implement Queue using Stacks

题目：

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 emptyoperations 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).

题解：

用一个辅助栈逆序，不过每push一个元素就要反转整个栈，非常耗时，不可取

Solution 1 

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

设两个栈，一个只能push，一个只能pop。有些情况需要注意，只有s2栈空了，才能把s1栈元素push进s2，而且一定要全部push进s2。另外在返回首元素时，有时候是刚开始push进s1，还没有pop操作（也就是说s1的元素还没有push进s2，此时s2为空，但是s1不为空），那么就要和pop操作一样要将栈1元素全部push进栈2。

Solution 2

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() {
        if(s2.empty() && s1.empty()){
            return INT_MIN;
        } else if(s2.empty()){
            while(!s1.empty()){
                int tmp = s1.top();
                s1.pop();
                s2.push(tmp);
            }
        }
        int tmp = s2.top();
        s2.pop();
        return tmp;
    }
    
    /** Get the front element. */
    int peek() {
        if(!s2.empty())
            return s2.top();
        while(!s1.empty()){
                int tmp = s1.top();
                s1.pop();
                s2.push(tmp);
        }
        return s2.top();
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return s1.empty() && s2.empty();
    }
    stack<int> s1, s2;
};