#-*- coding: UTF-8 -*- class Solution(object):    def removeElement(self, nums, val):        """        :type nums: List[int]        :type val: int        :rtype: int        """        for i in range(len(nums)-1,-1,-1):      …

#Method 1import math class Solution(object):    def majorityElement(self, nums):        numsDic={}        for num in nums:            numsDic[num]=numsDic[nums]+1 if num in numsDic else 0            if numsDic[num]>len(nums)/2:                return…

Given an array nums and a value val, remove all instances of that value in-placeand return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. The order of elem…

#-*- coding: UTF-8 -*- # Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def removeElements(self, head, val):        """      …

#-*- coding: UTF-8 -*-#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 = Noneclass Solution(object):    def re…

#-*- coding: UTF-8 -*- # Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def deleteDuplicates(self, head):        if head==None or head.…

#-*- coding: UTF-8 -*-class Solution(object):    def removeDuplicates(self, nums):        """        :type nums: List[int]        :rtype: int        """        if len(nums)<=1:return len(nums)        pre=0;next=1        wh…

1.题目 27. Remove Element——Easy Given an array nums and a value val, remove all instances of that value in-place and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1)…

Remove Element Given an array and a value, remove all instances of that value in place and return the new length. The order of elements can be changed. It doesn't matter what you leave beyond the new length. 解法一: 常规做法,设置一个新A数组的下标ind. 遍历过程中遇到elem就跳过,否…

#-*- 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.…