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; } }
