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LinkHash**特性：唯一和有序
由【哈希表】保证【键】的唯一性，由【链表】保证【键】的有序性(存储和取出的顺序一致)【LinkHashSet】和【LinkHashMap key】的特性是完全一样的！public class Test {public static void main(String[] args) {LinkedHashMap<Person, String> linkedHashMap = new LinkedHashMap<Person, String>();linkedHashMap.put(new Person("ea", 150), "包青天");linkedHashMap.put(new Person("ab", 27), "包青天");linkedHashMap.put(new Person("c", 150), "包青天");linkedHashMap.put(new Person("c", 160), "baiqiantao");linkedHashMap.put(new Person("ea", 160), "baiqiantao");linkedHashMap.put(new Person("d", 27), "包青天");System.out.println(linkedHashMap);//{ea,150=baiqiantao, ab,27=包青天, c,150=baiqiantao, d,27=包青天}}}class Person {public String name;public int age;public Person(String name, int age) {this.name = name;this.age = age;}public int hashCode() {int newAge = this.age;if (age > 150) newAge = 150;return name.hashCode() + newAge;//这里的hashCode是基于String类的hashCode更改的}public boolean equals(Object obj) {if (this == obj) return true;if (!(obj instanceof Person)) return false;Person p = (Person) obj;return this.name.equals(p.name);//只要name相同就认为equal值为true}public String toString() {//默认返回值是【person.getClass().getName() + ‘@‘ + Integer.toHexString(person.hashCode())】return name + "," + age;}}
Collection案例:模拟斗地主的牌局
public class Test {public static void main(String[] args) {// 创建一个牌盒List<PokerBean> list = new ArrayList<PokerBean>();String[] colors = { "?", "?", "?", "?" };String[] numbers = { "A", "2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "Q", "K" };// 装牌for (int i = 0; i < numbers.length; i++) {for (String color : colors) {list.add(new PokerBean(color, numbers[i], i + 1));}}list.add(new PokerBean("小王", "小王", 14));list.add(new PokerBean("大王", "小王", 15));// 洗牌Collections.shuffle(list);// 发牌Set<PokerBean> player_1 = new TreeSet<PokerBean>();Set<PokerBean> player_2 = new TreeSet<PokerBean>();Set<PokerBean> player_3 = new TreeSet<PokerBean>();Set<PokerBean> diPai = new TreeSet<PokerBean>();for (int x = 0; x < list.size(); x++) {if (x >= list.size() - 3) diPai.add(list.get(x));else if (x % 3 == 0) player_1.add(list.get(x));else if (x % 3 == 1) player_2.add(list.get(x));else if (x % 3 == 2) player_3.add(list.get(x));}//看牌System.out.println("玩家一的牌为：" + player_1);System.out.println("玩家二的牌为：" + player_2);System.out.println("地主家的牌为：" + player_3);System.out.println("底牌为：" + diPai);}}class PokerBean implements Comparable<PokerBean> {public String color;public String numberString;public int numberInt;public PokerBean(String color, String numberString, int numberInt) {this.color = color;this.numberString = numberString;this.numberInt = numberInt;}@Overridepublic String toString() {switch (numberInt) {case 10:return "[" + color + numberString + "]";//没有空格case 14:case 15:return "[" + numberString + "]";//没有空格，没有数字default:return "[ " + color + numberString + " ]";}}@Overridepublic int compareTo(PokerBean o) {int tem = this.numberInt - o.numberInt;return tem == 0 ? this.color.compareTo(o.color) : tem;}}
TreeMap案例:统计字符出现的次数
public class Test {public static void main(String[] args) {String str = "fd*+a//dA11 \n aDdD-d";System.out.println(getcharCount(str));}// 获取字符串中每一个字母出现的次数，要求打印结果形式为：a-2 \n b-1 形式public static String getcharCount(String str) {TreeMap<Character, Integer> map = new TreeMap<Character, Integer>();char[] arr = str.toCharArray();for (int i = 0; i < arr.length; i++) {if (!(arr[i] <= ‘z‘ && arr[i] >= ‘a‘ || arr[i] <= ‘Z‘ && arr[i] >= ‘A‘)) //限制查询范围continue;//表示不再执行后面的代码，继续下次遍历！Integer value = map.get(arr[i]);//Returns the value to which the specified key is mapped, or null if this map contains no mapping for the key.if (value == null) map.put(arr[i], 1);//若此字符不存在，则存入此字符，将值设为1else map.put(arr[i], value + 1);//若此字符已经存在，则值加1}return mapToString(map);}//将集合中的内容以指定格式的字符串形式返回private static <K, V> String mapToString(Map<K, V> map) {StringBuilder sb = new StringBuilder();Iterator<K> it = map.keySet().iterator();while (it.hasNext()) {K key = it.next();V value = map.get(key);sb.append(key + "-" + value + "\n");}return sb.toString();}}
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原文地址：http://www.cnblogs.com/baiqiantao/p/5449115.html
