• [LeetCode][JavaScript]Implement Stack using Queues


    Implement Stack using Queues

    Implement the following operations of a stack using queues.

    • push(x) -- Push element x onto stack.
    • pop() -- Removes the element on top of the stack.
    • top() -- Get the top element.
    • empty() -- Return whether the stack is empty.
    Notes:
    • You must use only standard operations of a queue -- which means only push to backpeek/pop from frontsize, and is empty operations are valid.
    • Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
    • You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

    Update (2015-06-11):
    The class name of the Java function had been updated to MyStack instead of Stack.

    https://leetcode.com/problems/implement-stack-using-queues/

     

     
     
     
    用队列模拟栈。
    只能用队列的入队,出队,长度和是否为空四个功能。
    也就是js数组的push(),shift(),length,length===0。
     
    两个队列倒来倒去。
    入栈操作就是找到不为空的队列,入队。
    出栈和查看栈顶元素的操作是一样的,把不为空的那个队列,除了最后一个元素的其他所有元素都倒到另一个里去。
    栈是否为空就是两个队列是否都为空。
     
     1 /**
     2  * @constructor
     3  */
     4 var Stack = function() {
     5     this.queueA = [];
     6     this.queueB = [];
     7 };
     8 
     9 /**
    10  * @param {number} x
    11  * @returns {void}
    12  */
    13 Stack.prototype.push = function(x) {
    14     if(this.queueA.length !== 0){
    15         this.queueA.push(x);
    16     }else{
    17         this.queueB.push(x);
    18     }
    19 };
    20 
    21 Stack.prototype.moveAToB = function() {
    22     if(this.queueB.length !== 0){
    23         var tmp = this.queueA;
    24         this.queueA = this.queueB;
    25         this.queueB = tmp;
    26     }
    27     var len = this.queueA.length;
    28     for(var i = 0; i < len - 1; i++){
    29         var queuePeek = this.queueA.shift();
    30         this.queueB.push(queuePeek);
    31     }
    32 };
    33 
    34 /**
    35  * @returns {void}
    36  */
    37 Stack.prototype.pop = function() {
    38     this.moveAToB();
    39     this.queueA.shift();
    40 };
    41 
    42 /**
    43  * @returns {number}
    44  */
    45 Stack.prototype.top = function() {
    46     this.moveAToB();
    47     var stackTop = this.queueA.shift();
    48     this.queueB.push(stackTop);
    49     return stackTop;
    50 };
    51 
    52 /**
    53  * @returns {boolean}
    54  */
    55 Stack.prototype.empty = function() {
    56     if(this.queueA.length === 0 && this.queueB.length === 0){
    57         return true;
    58     }
    59     return false;
    60 };
     
     
     
     
     
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  • 原文地址:https://www.cnblogs.com/Liok3187/p/4572856.html
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