day11 --------------------------------------------------------------- 实例018:复读机相加 题目 求s=a+aa+aaa+aaaa+aa-a的值,其中a是一个数字.例如2+22+222+2222+22222(此时共有5个数相加),几个数相加由键盘控制. 分析:很简单,字符串*x可以复制. 1 a = input('请输入数字:') 2 n = input("请输入要加几次:") 3 s = 0 4 for i in
算法导论:22页2.3-7 描述一个运行时间为O(nlogn)的算法,找出n个元素的S数组中是否存在两个元素相加等于给定x值 AC解: a=[1,3,6,7,9,15,29] def find2sumx(nums,x): nums.sort() le,ri=0,len(nums)-1 while le>=0 and ri<=len(nums) and le<ri: if nums[le]+nums[ri]<x: le+=1 elif nums[le]+nums[ri]>x:
# 已知有一个已经排好序的数组.要求是,有一个新数据项,要求按原来的规律将它插入数组中. a=[1,2,3,4,5,6,7,8,9]num=int(input("input num:"))for i in a: if i>num: ind=a.index(i) a.insert(ind,num) break else:continueprint(a)
Given an array of integers that is already sorted in ascending order, 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 m
代码: t = [-10,-3,-100,-1000,-239,1] # 交换 -10和1的位置 t[5], t[t[5]-1] = t[t[5]-1], t[5] 报错: IndexError: list assignment index out of range 数组: >>> t [-10,-3,-100,-1000,-239,-10] 为什么? 等式右边 t[t[5]-1] 相当于 t[0] ,是对值-10的引用.首先是将t[5]的引用指向-10,此时 t[5] 的值变为-10,
最近被算法虐了一下,刷一下leetcode,找找存在感 如题: Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Exam
package com.hanqi; import java.util.*; public class yonghukongzhi { public static void main(String[] args) { // TODO 自动生成的方法存根 System.out.println("请输入一个数:"); Scanner jishu = new Scanner(System.in); int a = jishu.nextInt(); System.out.println(&qu