#-*- coding: UTF-8 -*-class Stack(object):    def __init__(self):        """        initialize your data structure here.        """        self.inQueue=[]        self.outQueue=[]    def push(self, x):        """…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/implement-stack-using-queues/#/description 题目描述 Implement the following operations of a stack using queues. push(x) – Push element x onto…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 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…

题目: 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 on…

#-*- coding: UTF-8 -*- class MinStack(object):    def __init__(self):        """        initialize your data structure here.        """        self.Stack=[]        self.minStack=[]            def push(self, x):        "&…

用队列实现栈.这个实现方法十分的简单,就是在push这一步的时候直接变成逆序. class MyStack { private: queue<int> q; queue<int> q2; public: /** Initialize your data structure here. */ MyStack() { } /** Push element x onto stack. */ void push(int x) { q.push(x); ;i<q.size()-;i++…

#-*- coding: UTF-8 -*- #题意:大海捞刀,在长字符串中找出短字符串#AC源码:滑动窗口双指针的方法class Solution(object):    def strStr(self, hayStack, needle):        """        :type haystack: str        :type needle: str        :rtype: int        """        if…

#-*- coding: UTF-8 -*-#双栈法class Queue(object):    def __init__(self):        """        initialize your data structure here.        """        self.inStack=[]        self.outStack=[]            def push(self, x):        "…

155. Min Stack class MinStack { public: /** initialize your data structure here. */ MinStack() { } void push(int x) { if(s1.empty() && s2.empty()){ s1.push(x); s2.push(x); } else{ if(x < s2.top()){ s1.push(x); s2.push(x); } else{ s1.push(x); s2…

232. Implement Queue using Stacks Total Accepted: 27024 Total Submissions: 79793 Difficulty: Easy 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 o…

翻译 用队列来实现栈的例如以下操作. push(x) -- 将元素x加入进栈 pop() -- 从栈顶移除元素 top() -- 返回栈顶元素 empty() -- 返回栈是否为空 注意: 你必须使用一个仅仅有标准操作的队列. 也就是说,仅仅有push/pop/size/empty等操作是有效的. 队列可能不被原生支持.这取决于你所用的语言. 仅仅要你仅仅是用queue的标准操作,你能够用list或者deque(double-ended queue)来模拟队列. 你能够如果全部的操作都是有效的(…

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 may assume that…

#-*- coding: UTF-8 -*- #既然不能使用加法和减法,那么就用位操作.下面以计算5+4的例子说明如何用位操作实现加法:#1. 用二进制表示两个加数,a=5=0101,b=4=0100:#2. 用and(&)操作得到所有位上的进位carry=0100;#3. 用xor(^)操作找到a和b不同的位,赋值给a,a=0001:#4. 将进位carry左移一位,赋值给b,b=1000:#5. 循环直到进位carry为0,此时得到a=1001,即最后的sum.#!!!!!!关于负数的运算.…

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 s…

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 s…

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 s…

题目描述: 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 …

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 standa…

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. Example: MyStack stack = n…

#-*- coding: UTF-8 -*- #AC源码[意外惊喜,还以为会超时]class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """         for i in xrange(…

#-*- coding: UTF-8 -*-#利用strip函数去掉字符串去除空格(其实是去除两边[左边和右边]空格)#利用split分离字符串成列表class Solution(object):    def lengthOfLastWord(self, s):        """        :type s: str        :rtype: int        """        if s==None:return 0     …

#-*- coding: UTF-8 -*-#需要考虑多种情况#以下几种是可以返回的数值#1.以0开头的字符串,如01201215#2.以正负号开头的字符串,如'+121215':'-1215489'#3.1和2和空格混合形式[顺序只能是正负号-0,空格位置可以随意]的:'+00121515'#4.正数小于2147483647,负数大于-2147483648的数字#其他的情况都是返回0,因此在判断 是把上述可能出现的情况列出来,其他的返回0#AC源码如下class Solution(object…

用两个队列去实现栈,这里我使用了队列数组q[2],在所有的过程中保证一个队列是空的 push时插入到空的队列中,然后将队列中的元素移到另一个队列中 pop时从不空的队列中pop() peek时从不空的队列中取出front() class Stack { public: queue<]; // Push element x onto stack. void move(int x){ -x].empty()){ q[x].push(q[-x].front()); q[-x].pop(); } } v…

1.两个队列实现,始终保持一个队列为空即可 class MyStack { public: /** Initialize your data structure here. */ MyStack() { } /** Push element x onto stack. */ void push(int x) { if(que1.empty()) { que1.push(x); while(!que2.empty()) { int val=que2.front(); que2.pop(); que…

#-*- coding: UTF-8 -*-class Solution(object):    def compareVersion(self, version1, version2):        """        :type version1: str        :type version2: str        :rtype: int        """        versionl1=version1.split('.'…

class Solution(object):    def convertToTitle(self, n):        """        :type n: int        :rtype: str        """        res=''        while n>0:            tmp=n            n=(n-1)/26            res+=chr(65+(tmp-1)%26)…

#-*- coding: UTF-8 -*-#2147483648#在32位操作系统中,由于是二进制,#其能最大存储的数据是1111111111111111111111111111111.#正因为此,体现在windows或其他可视系统中的十进制应该为2147483647.#32位数的范围是 -2147483648~2147483648class Solution(object):    def reverse(self, x):        """        :type…

#-*- coding: UTF-8 -*-#由于题目要求不返回任何值,修改原始列表,#因此不能直接将新生成的结果赋值给nums,这样只是将变量指向新的列表,原列表并没有修改.#需要将新生成的结果赋予给nums[:],才能够修改原始列表class Solution(object):    def rotate(self, nums, k):        """        :type nums: List[int]        :type k: int        :…

#-*- coding: UTF-8 -*-# The isBadVersion API is already defined for you.# @param version, an integer# @return a bool# def isBadVersion(version):class Solution(object):    def firstBadVersion(self, n):        """        :type n: int        :…

#-*- coding: UTF-8 -*- class Solution(object):    def isPalindrome(self, s):        """        :type s: str        :rtype: bool        """        s=s.lower()        if s==None:return False        isPalindrome1=[]        isPal…