package day01; import java.util.Arrays; import java.util.Random; public class MaxOfArray { public static void main(String[] args) { int[] arr = new int[10]; Random ran = new Random(); //随机生成数 for(int i = 0;i<=9;i++) { arr[i] = ran.nextInt(100); } Sys…
网吧充值系统namespace ConsoleApplication1 { class Program { struct huiyuan { public string name; public string password; public double yue; } static void Main(string[] aaa) { ArrayList Ul = new ArrayList(); while (true) { try { Console.WriteLine("请输入您要执行的操…
package bianchengti; /* * 在由N个元素构成的集合S中,找出最小元素C,满足C=A-B, * 其中A,B是都集合S中的元素,没找到则返回-1 */ public class findMinValue { //快速排序 public static void sort(int a[], int low, int hight) { if (low > hight) { return; } int i, j, key; i = low; j = hight; key = a[i]…
现在有一个长度20的SET,其中每个对象的内容是随机生成的字符串,请写出遍历删除LIST里面字符串含"2"的对象的代码. public class RemoveTwo { //length用户要求产生字符串的长度 public static String getRS(int length){ String str="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"; Random rand…
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. 这里题目要求找出出现次数超过n/2的元素. 可以先排序,…
Given an array of integers, every element appears twice except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 数组中除了某个元素出现一次,其他都出现两次,找出只出现一次的元素. 一个数字和自己异或…
曾经看到有这样一个JS题:有一组数字,从1到n,从中减少了3个数,顺序也被打乱,放在一个n-3的数组里请找出丢失的数字,最好能有程序,最好算法比较快假设n=10000 下面我也来贴一个算法. function getArray (){ //创建随机丢失3个数字的数组,并打乱顺序. var arr =[] for(var i=1;i<=10000;i++){ arr.push(i); } var a = arr.splice(Math.floor(Math.random()*arr.length)…
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. Note: The solution set must not contain duplicate triplets. For example, given array S = [-1,…
就是找x+y=-z的组合 转化为找出值为-z满足x+y=-z的组合 解法一: 为了查找,首先想到排序,为了后面的二分,nlogn, 然后x+y的组合得n^2的复杂度,加上查找是否为-z,复杂度为nlogn + n^2 * logn 解法二: 还是先从小到大排序 nlogn 假设数组排序后为 a b c d e f 我们还是要找x+y=-z 会发现-z存在的可能只能是a+f和b+e,不会存在a+e和b+f这种情况(这里很重要,保证了算法的正确性),所以两个指针一头一尾往中间扫,肯定能找出来 fis…