浅谈一下Java中集合的迭代方式

		List<Dog> dogs = new ArrayList<>();
		dogs.add(new Dog("大黄", 2));
		dogs.add(new Dog("小黑", 3));
		dogs.add(new Dog("小哈",1));
		
		//最原始的 for 循环
		for (int i = 0; i < dogs.size(); i++) {
			System.out.println(dogs.get(i));
		}
		//增强 for 循环
		for (Dog dog1 : dogs) {
			System.out.println(dog1);
		}
		//使用迭代器
		Iterator<Dog> it = dogs.iterator();
		while (it.hasNext()) {
			Dog dog2 = it.next();
			System.out.println(dog2);
		}

//使用 Stream 和 Lambda 表达式进行迭代
dogs.stream().forEach(System.out::println);

//只使用 Lambda 表达式进行迭代
dogs.forEach(System.out::println);

List<String> mixStr = new ArrayList<String>(Arrays.asList("1", "2", "3", "a", "4", "5", "c", "6", "7", "8", "d", "9", "10"));
int sumary = mixStr.
		stream().
		filter(s->s.matches("^[0-9]*$")).
		map(Integer::parseInt).
		reduce(0, (acc, e)->acc+e);
System.out.println("方式一（推荐）："+sumary);

//有时候，一行太长了，可以分开写，这样增加了可读性。
sumary = mixStr.stream().filter(s->s.matches("^[0-9]*$")).map(Integer::parseInt).reduce(0, (acc, e)->acc+e);

Stream<String> filter = mixStr.stream().filter(s->s.matches("^[0-9]*$"));
Stream<Integer> map = filter.map(Integer::parseInt);
Integer reduce = map.reduce(0, (acc, e)->acc+e);
System.out.println(""+reduce);
System.out.println("方式二（不推荐）："+reduce);

sumary = 0;
for (String s : mixStr) {
	if (s.matches("^[0-9]*$")) {
		sumary += Integer.parseInt(s);
	}
}
System.out.println("传统的方式："+sumary);

//流的用处很多。
mixStr.stream().filter(s->s.matches("^[0-9]*$")).map(Integer::parseInt).forEach(System.out::print);