Mryang2333的博客

7.整数反转（通过）给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转class Solution: def reverse(self, x): """ :type x: int :rtype: int """ y = abs(x) l = list(str(y)) ..

首先我想到的是直接return x**0.5但是人家明显不让这么做。。。然后借鉴。。。牛顿迭代法实现。class Solution: def mySqrt(self, x): """ :type x: int :rtype: int """ if x <= 1: retur...

前9阶的解法分别是：1,2,3,5,8,13,21,34,55由上可知，与Fibonacci sequence及其相似class Solution: def climbStairs(self, n): """ :type n: int :rtype: int """ tmp = 1 res ...

和c有点不一样，Python不需要释放被删除的节点# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: def deleteDuplicate...

上学那会写的最多的就是前中后层遍历，各种转换。。。# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solut...

用的递归在第100题的基础上改一改就可以了# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution...

二叉树的最大深度# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def maxD...

class Solution: def titleToNumber(self, s): """ :type s: str :rtype: int """ num = 0 for i in range(len(s)-1,-1,-1): num += (ord(s[i]) -...

最开始想用s.reverse()后来还是写了个替换class Solution: def reverseString(self, s): """ :type s: str :rtype: str """ l = len(s) - 1 for i in range(l): ..

class Solution: def hammingDistance(self, x, y): """ :type x: int :type y: int :rtype: int """ count = 0 while True: if x == 0 a...

二进制求和最蠢写法（没有用bin()和int()）后边附上使用int和bin版本class Solution: def addBinary(self, a, b): """ :type a: str :type b: str :rtype: str """ a = list(map(int,l...

map()函数Python 2.x 返回列表Python 3.x 返回迭代器。python2class Solution(object): def plusOne(self, digits): """ :type digits: List[int] :rtype: List[int] """ digit...

判断一个整数是否是回文数。回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。class Solution: def isPalindrome(self, x): """ :type x: int :rtype: bool """ y = list(str

class Solution: def romanToInt(self, s): """ :type s: str :rtype: int """ y = list(s) vl = {'I':1,'V':5,'X':10,'L':50,'C':100,

给定一个只包括 {} 的字符串，判断字符串是否有效。class Solution: def isValid(self, s): """ :type s: str :rtype: bool """ vl = {'{':1,'}':-1,'[':2,']':-2,'(':3,')':-3} y ..

emmm…这个答案有问题，太慢了，不让用set（），难受class Solution: def removeDuplicates(self, nums): """ :type nums: List[int] :rtype: int """ count = 0 for i in range(le..

class Solution: def removeElement(self, nums, val): """ :type nums: List[int] :type val: int :rtype: int """ count = nums.coun

偷懒写法，emmm看了看题名，想了一下，这应该是让我实现Python的index（）吧 def removeElement(self, nums, val): """ :type nums: List[int] :type val: int :rtype: int """ try: hay..

给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。你可以假设数组中无重复元素。class Solution: def searchInsert(self, nums, target): """ :type nums: List[int] :type target: in...

class Solution: def maxSubArray(self, nums): """ :type nums: List[int] :rtype: int """ sum = 0 Maxsum = nums[0] for i in range(len(nums)):...

给定一个仅包含大小写字母和空格’ ’ 的字符串，返回其最后一个单词的长度。如果不存在最后一个单词，请返回 0说明：一个单词是指由字母组成，但不包含任何空格的字符串。class Solution: def lengthOfLastWord(self, s): """ :type s: str :rtype: int ""...

class Solution: def arrayPairSum(self, nums): """ :type nums: List[int] :rtype: int """ listNum = list(nums) listNum.sort() sum = 0 ...