a=[1,2,3] b=[4,5,6] 现将list a与 list b按位相加,其结果为[5,7,9] 方法一: c=[a[i]+b[i] for i in range(min(len(a),len(b)))] 方法二: c=list(map(lambda x :x[0]+x[1] ,zip(a,b))) 方法三: 调用numpy库 import numpy as np c = np.array(a) + np.array(b) map()函数: map()函数接受两个参数,一个是函数,一个是
题目: You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -&
题目链接:https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/#/description 给定一个已经排好序的数组,数组中元素存在重复,如果允许一个元素最多可出现两次,求出剔除重复元素(出现次数是两次以上的)后的数组的长度. For example, Given sorted array nums = [1,1,1,2,2,3], Your function should return length = 5,
题目: Given an array of integers, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note t
分析:长整数相加,将结果分为高位和低位部分,分别保存在两个32整数中. 比如:unsigned int a = 0xFFFFFFFF, unsigned int b = 0x1, 结果用unsigned int c保存,c = a + b ,这样c的结果是0x00000000,因为高于32位的部分被截断了,所以 低位部分的结果就是c里保存的内容,再用一个unsigned int变量保存结果的高位部分,高位部分只可能有两种值, 0 或 1 , 就好比十进制两个一位数相加,最大也就是9+9 , 进位
转自:http://www.maomao365.com/?p=7205 摘要: 下文分享两条sql求和脚本,再次求和的方法分享 /* 例: 下文已知两条sql求和脚本,现需对两张不同表的求和记录再次求和 */ ---对两条求和sql脚本求和的方法 select sum(q) from ( select sum(qty) as q from tableNameA where ... union all select sum(qty) as q from tableNameB where ... )