#-*- coding: UTF-8 -*- class Solution(object):    hexDic={0:'0',1:'1',2:'2',3:'3',4:'4',5:'5',6:'6',7:'7',8:'8',9:'9',\            10:'a',            11:'b',            12:'c',            13:'d',            14:'e',            15:'f'}            def t…

#-*- coding: UTF-8 -*-# The guess API is already defined for you.# @param num, your guess# @return -1 if my number is lower, 1 if my number is higher, otherwise return 0# def guess(num):#binary searchclass Solution(object):    def guessNumber(self, n…

#回文数#Method1:将整数转置和原数比较,一样就是回文数:负数不是回文数#这里反转整数时不需要考虑溢出,但不代表如果是C/C++等语言也不需要考虑class Solution(object):    def isPalindrome(self, x):        """        :type x: int        :rtype: bool        """        if x<0:return False    …

class Solution(object):    def isUgly(self, num):        if num<=0:return False        comlist=[2,3,5];modflag=0                while True:            if num==1:break            for value in comlist:                tuple=divmod(num,value)            …

#-*- 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.…

#-*- 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 -*- class Solution(object):    def hammingWeight(self, n):        if n<=0:return n        mid=[]        while True:            if n==0:break            n,mod=divmod(n,2)            mid.append(mod)        mid.reverse()        return…

#-*- coding: UTF-8 -*- # ord(c) -> integer##Return the integer ordinal of a one-character string.##参数是一个ascii字符,返回值是对应的十进制整数class Solution(object):    def titleToNumber(self, s):        columns=0        n=len(s)        s=s.upper()[::-1]        for c…

Problem Link: http://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/ We design a auxilar function that convert a linked list to a node with following properties: The node is the mid-node of the linked list. The node's left child 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(…