## 题目地址 https://leetcode.com/problems/implement-queue-using-stacks/description/ ## 题目描述 ``` 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. Example: MyQueue queue = new MyQueue(); queue.push(1); queue.push(2); queue.peek(); // returns 1 queue.pop(); // returns 1 queue.empty(); // returns false 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). ``` ## 思路 这道题目是让我们用栈来模拟实现队列。 我们直到栈和队列都是一种受限的数据结构。 栈的特点是只能在一端进行所有操作,队列的特点是只能在一端入队,另一端出队。 在这里我们可以借助另外一个栈,也就是说用两个栈来实现队列的效果。这种做法的时间复杂度和空间复杂度都是O(n)。 由于栈只能操作一端,因此我们peek或者pop的时候也只去操作顶部元素,要达到目的 我们需要在push的时候将队头的元素放到栈顶即可。 因此我们只需要在push的时候,用一下辅助栈即可。 具体做法是先将栈清空并依次放到另一个辅助栈中,辅助栈中的元素再次放回栈中,最后将新的元素push进去即可。 比如我们现在栈中已经是1,2,3,4了。 我们现在要push一个5. push之前是这样的: ![232.implement-queue-using-stacks.drawio](../assets/problems/232.implement-queue-using-stacks-1.jpg) 然后我们将栈中的元素转移到辅助栈: ![232.implement-queue-using-stacks.drawio](../assets/problems/232.implement-queue-using-stacks-2.jpg) 最后将新的元素添加到栈顶。 ![232.implement-queue-using-stacks.drawio](../assets/problems/232.implement-queue-using-stacks-3.jpg) 整个过程是这样的: ![232.implement-queue-using-stacks.drawio](../assets/problems/232.implement-queue-using-stacks-4.jpg) ## 关键点解析 - 在push的时候利用辅助栈(双栈) ## 代码 * 语言支持:JS, Python Javascript Code: ```js /* * @lc app=leetcode id=232 lang=javascript * * [232] Implement Queue using Stacks */ /** * Initialize your data structure here. */ var MyQueue = function() { // tag: queue stack array this.stack = []; this.helperStack = []; }; /** * Push element x to the back of queue. * @param {number} x * @return {void} */ MyQueue.prototype.push = function(x) { let cur = null; while ((cur = this.stack.pop())) { this.helperStack.push(cur); } this.helperStack.push(x); while ((cur = this.helperStack.pop())) { this.stack.push(cur); } }; /** * Removes the element from in front of queue and returns that element. * @return {number} */ MyQueue.prototype.pop = function() { return this.stack.pop(); }; /** * Get the front element. * @return {number} */ MyQueue.prototype.peek = function() { return this.stack[this.stack.length - 1]; }; /** * Returns whether the queue is empty. * @return {boolean} */ MyQueue.prototype.empty = function() { return this.stack.length === 0; }; /** * Your MyQueue object will be instantiated and called as such: * var obj = new MyQueue() * obj.push(x) * var param_2 = obj.pop() * var param_3 = obj.peek() * var param_4 = obj.empty() */ ``` Python Code: ```python class MyQueue: def __init__(self): """ Initialize your data structure here. """ self.stack = [] self.help_stack = [] def push(self, x: int) -> None: """ Push element x to the back of queue. """ while self.stack: self.help_stack.append(self.stack.pop()) self.help_stack.append(x) while self.help_stack: self.stack.append(self.help_stack.pop()) def pop(self) -> int: """ Removes the element from in front of queue and returns that element. """ return self.stack.pop() def peek(self) -> int: """ Get the front element. """ return self.stack[-1] def empty(self) -> bool: """ Returns whether the queue is empty. """ return not bool(self.stack) # Your MyQueue object will be instantiated and called as such: # obj = MyQueue() # obj.push(x) # param_2 = obj.pop() # param_3 = obj.peek() # param_4 = obj.empty() ``` ## 扩展 - 类似的题目有用队列实现栈,思路是完全一样的,大家有兴趣可以试一下。 - 栈混洗也是借助另外一个栈来完成的,从这点来看,两者有相似之处。