求两个数组的交集并集并对结果排序

   这个是在自己学习了容器之后写的 

 package week13container;/*对week13container.IntersectJiaoJi2.java的改进 * */import java.util.*;public class IntersectJiaoJi3 {public static void main(String args[]) {int[] a = { 4, 6, 7, 7, 7, 7, 8, 8, 9, 10, 100, 130, 130, 140, 150 };int[] b = { 2, 3, 4, 4, 4, 4, 7, 8, 8, 8, 8, 9, 100, 130, 150, 160 };intersection(a, b);}static void intersection(int a[], int b[]) {Set<Integer> set1 = new HashSet<Integer>();Set<Integer> set2 = new HashSet<Integer>();// 去除掉数组a和b中重复的元素for (int i = 0; i < a.length; i++) {set1.add(a[i]);}for (int j = 0; j < b.length; j++) {set2.add(b[j]);}// 交集Set<Integer> su = new HashSet<Integer>(set1);su.retainAll(set2);System.out.println("两个数组的交集:" + sortSet(su) + "/n交集元素个数为" + su.size());// 并集Set<Integer> sn = new HashSet<Integer>(set1);sn.addAll(set2);System.out.println("两个数组的并集:" + sortSet(sn) + "/n并集元素个数为" + sn.size());}static String sortSet(Set<Integer> set) {//对Set进行排序的方法Integer setArray[] = new Integer[set.size()];Iterator<Integer> iSet = set.iterator();for (int i = 0; iSet.hasNext(); i++) {setArray[i] = iSet.next();// 将无序的HashSet存入到有序的数组中}Arrays.sort(setArray);// 排序return Arrays.toString(setArray);//这句话执行的结果是String,所以方法的返回类型必须是String}}

这个是火龙果写的

package week11;/*阿里巴巴面试题 * 问题1：马尔科夫（HMM）的特征是什么？ *问题2：有两个有序整数集合a和b，写一个函数找出它们的交集？http://topic.csdn.net/u/20081012/14/3cc93688-1f7f-4985-806c-3f729c78261b.html * */import java.util.Arrays;public class IntersectJiaoJi {public static void main(String args[]) {int[] b = { 4, 6, 7, 7, 7, 7, 8, 8, 9, 10, 100, 130, 130, 140, 150 };int[] a = { 2, 3, 4, 4, 4, 4, 7, 8, 8, 8, 8, 9, 100, 130, 150, 160 };int[] c = intersect(a, b);System.out.println(Arrays.toString(c));}public static int[] intersect(int[] a, int[] b) {if (a[0] > b[b.length - 1] || b[0] > a[a.length - 1]) {return new int[0];}int[] intersection = new int[Math.max(a.length, b.length)];int offset = 0;for (int i = 0, s = i; i < a.length && s < b.length; i++) {//这个句子很有意思while (a[i] > b[s]) {s++;}if (a[i] == b[s]) {intersection[offset++] = b[s++];}while (i < (a.length - 1) && a[i] == a[i + 1]) {i++;}}if (intersection.length == offset) {return intersection;}int[] duplicate = new int[offset];System.arraycopy(intersection, 0, duplicate, 0, offset);return duplicate;}}