#-*- coding: UTF-8 -*-#遍历所有元素,将元素值当做键.元素下标当做值#存放在一个字典中.遍历的时候,#如果发现重复元素,则比较其下标的差值是否小于k,#如果小于则可直接返回True,否则更新字典中该键的值为新的下标class Solution(object):    def containsNearbyDuplicate(self, nums, k):        """        :type nums: List[int]        :typ…

#-*- coding: UTF-8 -*- class Solution(object):    def containsDuplicate(self, nums):        numsdic={}        for num in nums:            numsdic[num]=numsdic[num]+1 if num in numsdic else 1            if numsdic[num]>=2:                return True  …

#-*- 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.…

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 使用set 使用字典 日期 题目地址:https://leetcode.com/problems/contains-duplicate-ii/description/ 题目描述 Given an array of integers and an integer k, find out whether there are two distinct in…

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 Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j]and the differen…

题目: Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and jis at most k. 提示: 此题考察的是对Hash Table的应用.这里给出两种方法,第一种基于unordered_map(Has…

#-*- coding: UTF-8 -*- class Solution(object):    def intersect(self, nums1, nums2):                if len(nums1)<len(nums2):            tmp=nums1            nums1=nums2            nums2=tmp               tmpdic={}               for num in nums1:    …

#-*- coding: UTF-8 -*-#杨辉三角返回给定行#方法:自上而下考虑问题,从给定的一行计算出下一行,则给定行数之后,计算的最后一行就是求解的最后一行class Solution(object):    def getRow(self, rowIndex):        """        :type rowIndex: int        :rtype: List[int]        """        #初始化杨辉三…

#-*- coding: UTF-8 -*- # Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def levelOrderBottom(self, root):   …

https://leetcode.com/problems/palindrome-partitioning-ii/description/ [题意] 给定一个字符串,求最少切割多少下,使得切割后的每个子串都是回文串 [思路] 求一个字符串的所有子串是否是回文串,O(n^2) dp从后往前推 vector<vector<bool> > p(len,vector<bool>(len)); ;i<len-;i++){ ;j<len-;j++){ if(j!=i)…

#-*- 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…

Problem Link: http://oj.leetcode.com/problems/pascals-triangle-ii/ Let T[i][j] be the j-th element of the i-th row in the triangle, for 0 <= j <= i, i = 0, 1, ... And we have the recursive function for T[i][j] T[i][j] = 1, if j == 0 or j == i T[i][j…

Problem link: http://oj.leetcode.com/problems/linked-list-cycle-ii/ The solution has two step: Detecting the loop using faster/slower pointers. Finding the begining of the loop. After two pointers meet in step 1, keep one pointer and set anther to th…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space?   Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Fol…

Reverse Linked List II Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note:Given m, n satisfy the following cond…

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

#-*- 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…

#-*- coding: UTF-8 -*- #ZigZag Conversion :之字型class Solution(object):    def convert(self, s, numRows):        """        :type s: str        :type numRows: int        :rtype: str        """        if numRows==1:return s     …

#-*- coding: UTF-8 -*- #Tags:dynamic programming,sumRange(i,j)=sum(j)-sum(i-1)class NumArray(object):    sums=[]    def __init__(self, nums):        """        initialize your data structure here.        :type nums: List[int]        "&…

#-*- coding: UTF-8 -*- #Hint1:#数字i,i的倍数一定不是质数,因此去掉i的倍数,例如5,5*1,5*2,5*3,5*4,5*5都不是质数,应该去掉#5*1,5*2,5*3,5*4 在数字1,2,3,4的时候都已经剔除过,因此数字5,应该从5*5开始#Hint2:#例:#2 × 6 = 12#3 × 4 = 12#4 × 3 = 12#6 × 2 = 12#显然4 × 3 = 12,和6 × 2 = 12不应该分析,因为在前两式中已经知道12不是质数,因此如果数字i是…

#-*- coding: UTF-8 -*- #l1 = ['1','3','2','3','2','1','1']#l2 = sorted(sorted(set(l1),key=l1.index,reverse=False),reverse=True)class Solution(object):    def thirdMax(self, nums):        """        :type nums: List[int]        :rtype: int  …

#-*- 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 -*- class Solution:    # @param n, an integer    # @return an integer    def reverseBits(self, n):        tmp=str(bin(n))[2:][::-1]                for i in xrange(32-len(tmp)):            tmp+='0'               return int(tmp,base=…