轩辕互动-1.求三个整数中第二大的数2.整型数组的平衡点
import java.util.ArrayList;import java.util.Arrays;import java.util.List;public class ExoWeb {public static void main(String[] args) {ExoWeb ew=new ExoWeb();System.out.println(ew.findSecondIn3Item(2, 5, 3));int[][] aa={{1,1,1,1,2,1},{1, 1, 0, 0, 1, 1},{1,2,0,2,1},{1,-1,1},{1,8,7,8,25,4,49,30,23},{1, 3, 5, 7, 8, 25, 4, 20},};for(int[] each:aa){System.out.print(Arrays.toString(each)+",balance points=");List<Integer> points=ew.findBalancePoint(each);for(int i:points){System.out.print(i+",");}System.out.println();}}/* * find the mid item in THREE integers */public int findSecondIn3Item(int a,int b,int c){//could this be faster than sorting?int max=(a>b?a:b)>c?(a>b?a:b):c;int min=(a<b?a:b)<c?(a<b?a:b):c;int mid=a^b^c^max^min;//equals to a+b+c-max-min/*int[] x=new int[3];x[0]=a;x[1]=b;x[2]=c;for(int i=0;i<x.length;i++){for(int j=0;j<x.length-i-1;j++){if(x[j]>x[j+1]){int temp=x[j];x[j]=x[j+1];x[j+1]=temp;}}}int mid=x[1];*/return mid;}/* * 平衡点：比如int[] numbers = {1,3,5,7,8,25,4,20}; 25前面的总和为24，25后面的总和也是24，25这个点就是平衡点； * 假如一个数组中的元素，其前面的部分等于后面的部分，那么这个点的位序就是平衡点 * 要求1：返回任何一个平衡点 * 要求2：要求输入的数组可能是GB级 * solution to 1. use List<Integer> * solution to 2. i cannot figure out a elegant solution. GB,,so you just cann't sum all the data in array */public List<Integer> findBalancePoint(int[] a){List<Integer> points=new ArrayList<Integer>();int total=0;int firstSum=0;for(int i=0,len=a.length;i<len;i++){//若输入的数组是GB级，全部数组求和不可取total+=a[i];}for(int i=0,len=a.length;i<len;i++){if(firstSum==total-a[i]-firstSum){points.add(a[i]);}firstSum+=a[i];}return points;}}