一.程序编写 import java.util.*;public class Port { public static void main(String[] args) { // TODO 自动生成的方法存根 int a[]=new int[10]; Scanner d=new Scanner(System.in); System.out.println("输入十个整数:"); for(int i= 0; i < a.length; i++) a[i]=d.nex
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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 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,
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)
源代码: package 数组;import java.util.*;public class vvv { public static void main(String[] args) { Scanner s = new Scanner(System.in); int[] x = new int[10]; System.out.println("请输入长度为10的数组:"); for (int i = 0; i < 10; i++) { x[i] = s.nextInt(); }