【leetcode?python】Intersection of Two Arrays

#-*- 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

双指针思想:两个指针相隔n-1,每次两个指针向后一步,当后面一个指针没有后继了,前面一个指针的后继就是要删除的节点 # Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = None class Solution(object):    def removeNthFromEnd(self,

数独规则如下:相当于一个9*9的矩阵 代码如下:#特定的九个格内1-9的个数至多为1#依次检查每行,每列,每个子九宫格是否出现重复元素,如果出现返回false,否则返回true.class Solution(object):    def isValidSudoku(self, board):        """        :type board: List[List[str]]        :rtype: bool        """

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = None class Solution(object):    depthList=[]    def minDepth(self, root):        ""&q

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = None class Solution(object):    def removeElements(self, head, val):        """        :type head: ListNode

class Solution(object):    def merge(self, nums1, m, nums2, n):        """        :type nums1: List[int]        :type m: int        :type nums2: List[int]        :type n: int        :rtype: void Do not return anything, modify nums1 in-place

#很简单,利用strip函数去除左右两边的空格,然后用split函数分割成列表 class Solution(object):    def lengthOfLastWord(self, s):        """        :type s: str        :rtype: int        """        if s==None:return 0        slist=s.strip().split(' ')      

#栈是先进后出 #队列先进先出 class Stack(object):    def __init__(self):        """        initialize your data structure here.        """        self.inQueue=[]        self.outQueue=[] def push(self, x):        """       

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