一天一道LeetCode

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（一）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 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)

（二）解题

题目大意：用两个栈实现queue，要求实现push，pop，peek和empty
解题思路：栈是先进后出，队列是先进先出。利用两个栈stack1和stack2
push：和栈一样，直接push到一个栈stack1中
pop：有一个辅助栈就可以将stack1中的数依次push到stack2中，这样stack2的top就是最先进来的数，直接pop即可
peek：去队首的数，如果stack2不为空的话，直接返回stack2.pop即可，如果stack2为空，则先把stack1的数push过来，再去top元素
empty：如果stack1和stack2均为空，就直接返回true，否则返回false
具体实现如下：

class Queue {
public:
    // Push element x to the back of queue.
    void push(int x) {
        st1.push(x);
    }

    // Removes the element from in front of queue.
    void pop(void) {
        if(empty()) return;
        if(st2.empty())
        {
            while(!st1.empty()){//为空的话要把stack1的元素push进来
                st2.push(st1.top());
                st1.pop();
            }
        }
        st2.pop();
    }

    // Get the front element.
    int peek(void) {
        if(st2.empty()){//为空的话要把stack1的元素push进来
            while(!st1.empty()){
                st2.push(st1.top());
                st1.pop();
            }
        }
        return st2.top();
    }

    // Return whether the queue is empty.
    bool empty(void) {
        if(st1.empty()&&st2.empty()) return true;
        else return false;
    }
private:
    stack<int> st1;
    stack<int> st2;
};