标签：链表 图解 entry concat collect cte arc index ring
B:案例演示
Map集合的遍历之键找值
HashMap<String, Integer> hm = new HashMap<>();
hm.put("张三", 23);
hm.put("李四", 24);
hm.put("王五", 25);
hm.put("赵六", 26);
/*Set<String> keySet = hm.keySet(); //获取集合中所有的键
Iterator<String> it = keySet.iterator(); //获取迭代器
while(it.hasNext()) { //判断单列集合中是否有元素
String key = it.next(); //获取集合中的每一个元素,其实就是双列集合中的键
Integer value = hm.get(key); //根据键获取值
System.out.println(key + "=" + value); //打印键值对
}*/
for(String key : hm.keySet()) { //增强for循环迭代双列集合第一种方式
System.out.println(key + "=" + hm.get(key));
}
B:案例演示
Map集合的遍历之键值对对象找键和值
HashMap<String, Integer> hm = new HashMap<>();
hm.put("张三", 23);
hm.put("李四", 24);
hm.put("王五", 25);
hm.put("赵六", 26);
/*Set<Map.Entry<String, Integer>> entrySet = hm.entrySet(); //获取所有的键值对象的集合
Iterator<Entry<String, Integer>> it = entrySet.iterator();//获取迭代器
while(it.hasNext()) {
Entry<String, Integer> en = it.next(); //获取键值对对象
String key = en.getKey(); //根据键值对对象获取键
Integer value = en.getValue(); //根据键值对对象获取值
System.out.println(key + "=" + value);
}*/
for(Entry<String,Integer> en : hm.entrySet()) {
System.out.println(en.getKey() + "=" + en.getValue());
}
C:源码分析
A:案例演示
需求：统计字符串中每个字符出现的次数 String str = "aaaabbbcccccccccc"; char[] arr = str.toCharArray(); //将字符串转换成字符数组 HashMap<Character, Integer> hm = new HashMap<>(); //创建双列集合存储键和值
for(char c : arr) { //遍历字符数组
/*if(!hm.containsKey(c)) { //如果不包含这个键
hm.put(c, 1); //就将键和值为1添加
}else { //如果包含这个键
hm.put(c, hm.get(c) + 1); //就将键和值再加1添加进来
}
//hm.put(c, !hm.containsKey(c) ? 1 : hm.get(c) + 1);
Integer i = !hm.containsKey(c) ? hm.put(c, 1) : hm.put(c, hm.get(c) + 1);
}
for (Character key : hm.keySet()) { //遍历双列集合
System.out.println(key + "=" + hm.get(key));
}
public static <T> void sort(List<T> list)
public static <T> int binarySearch(List<?> list,T key)
public static <T> T max(Collection<?> coll)
public static void reverse(List<?> list)
public static void shuffle(List<?> list)
A:案例演示
模拟斗地主洗牌和发牌，牌没有排序
//买一副扑克
String[] num = {"A","2","3","4","5","6","7","8","9","10","J","Q","K"};
String[] color = {"方片","梅花","红桃","黑桃"};
ArrayList<String> poker = new ArrayList<>();
for(String s1 : color) {
for(String s2 : num) {
poker.add(s1.concat(s2));
}
}
poker.add("小王");
poker.add("大王");
//洗牌
Collections.shuffle(poker);
//发牌
ArrayList<String> gaojin = new ArrayList<>();
ArrayList<String> longwu = new ArrayList<>();
ArrayList<String> me = new ArrayList<>();
ArrayList<String> dipai = new ArrayList<>();
for(int i = 0; i < poker.size(); i++) {
if(i >= poker.size() - 3) {
dipai.add(poker.get(i));
}else if(i % 3 == 0) {
gaojin.add(poker.get(i));
}else if(i % 3 == 1) {
longwu.add(poker.get(i));
}else {
me.add(poker.get(i));
}
}
//看牌
System.out.println(gaojin);
System.out.println(longwu);
System.out.println(me);
System.out.println(dipai);
//买一副牌
String[] num = {"3","4","5","6","7","8","9","10","J","Q","K","A","2"};
String[] color = {"方片","梅花","红桃","黑桃"};
HashMap<Integer, String> hm = new HashMap<>(); //存储索引和扑克牌
ArrayList<Integer> list = new ArrayList<>(); //存储索引
int index = 0; //索引的开始值
for(String s1 : num) {
for(String s2 : color) {
hm.put(index, s2.concat(s1)); //将索引和扑克牌添加到HashMap中
list.add(index); //将索引添加到ArrayList集合中
index++;
}
}
hm.put(index, "小王");
list.add(index);
index++;
hm.put(index, "大王");
list.add(index);
//洗牌
Collections.shuffle(list);
//发牌
TreeSet<Integer> gaojin = new TreeSet<>();
TreeSet<Integer> longwu = new TreeSet<>();
TreeSet<Integer> me = new TreeSet<>();
TreeSet<Integer> dipai = new TreeSet<>();
for(int i = 0; i < list.size(); i++) {
if(i >= list.size() - 3) {
dipai.add(list.get(i)); //将list集合中的索引添加到TreeSet集合中会自动排序
}else if(i % 3 == 0) {
gaojin.add(list.get(i));
}else if(i % 3 == 1) {
longwu.add(list.get(i));
}else {
me.add(list.get(i));
}
}
//看牌
lookPoker("高进", gaojin, hm);
lookPoker("龙五", longwu, hm);
lookPoker("冯佳", me, hm);
lookPoker("底牌", dipai, hm);
}
public static void lookPoker(String name,TreeSet<Integer> ts,HashMap<Integer, String> hm) {
System.out.print(name + "的牌是:");
for (Integer index : ts) {
System.out.print(hm.get(index) + " ");
}
System.out.println();
}
标签：链表 图解 entry concat collect cte arc index ring
原文地址：http://www.cnblogs.com/GJ-ios/p/5998977.html
