关于Set和Map的注意事项

1. HashSet和HashMap

public boolean add(E e)

Adds the specified element to this set if it is not already present. More formally, adds the specified element e to this set if this set contains no element e2 such that (e==null ? e2==null : e.equals(e2)). If this set already contains the element, the call leaves the set unchanged and returns false.

public V put(K key, V value)

Associates the specified value with the specified key in this map. If the map previously contained a mapping for the key, the old value is replaced.

set add如果是相同的那么忽略新的，老的保持不变

map put如果是相同的key，那么更新老的value

refer http://stackoverflow.com/questions/12940663/does-adding-a-duplicate-value-to-a-hashset-hashmap-replace-the-previous-value

hash类型的set和map判断相同的标准是equals，但是也可能为了优化性能先用hashCode进行了过滤，因为如果equals是true的话，要保证这两个object的hashCode是相等的

refer

http://stackoverflow.com/questions/5396939/hashcode-and-equals-for-hashset

http://stackoverflow.com/questions/364794/what-does-java-use-to-determine-if-a-key-is-a-duplicate-in-a-map

默认情况下，new出来的对象都是不equals的，hashCode也是不相等的，实际上就是和创建出来的对象内存地址相关的

equals:

The equals method for class Object implements the most discriminating possible equivalence relation on objects; that is, for any non-null reference values x and y, this method returns true if and only if x and y refer to the same object (x == y has the value true).

hashCode:

As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects. (This is typically implemented by converting the internal address of the object into an integer, but this implementation technique is not required by the JavaTM programming language.)

refer http://stackoverflow.com/questions/27581/what-issues-should-be-considered-when-overriding-equals-and-hashcode-in-java

http://stackoverflow.com/questions/4178997/how-default-equals-and-hashcode-will-work-for-my-classes

所以，如果想要自定义一个对象的比较方式需要手动同时实现equals和hashCode

例如，如下的AppInfo希望set里面保留的都是id不同的对象

public static void main(String[] args) {
    AppInfo app1 = new AppInfo();
    app1.id = "id-1";
    app1.totalCount = 1;

    AppInfo app2 = new AppInfo();
    app2.id = "id-2";
    app2.totalCount = 2;

    AppInfo app3 = new AppInfo();
    app3.id = "id-1";
    app3.totalCount = 3;

    Set<AppInfo> set = new HashSet<>();
    set.add(app1);
    set.add(app2);
    set.add(app3);

    for (AppInfo app : set) {
        System.out.println(String.format("[%s, %d]", app.id, app.totalCount));
    }
}

static class AppInfo {
    public String id;
    public int totalCount;

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        AppInfo appInfo = (AppInfo) o;

        return id.equals(appInfo.id);
    }

    @Override
    public int hashCode() {
        return id.hashCode();
    }
}

结果是

[id-2, 2]

[id-1, 1]

2. TreeSet和TreeMap

TreeSet的内部元素是有序的，TreeMap的内部元素 key是有序的

这两种类型的数据结构并不会关注equals，而是直接通过Comparator的compare结果，或者是实现Comparable的compareTo结果来判定是否是重复的元素

refer http://stackoverflow.com/questions/28779766/use-of-hashcode-and-equals-in-treeset-and-treemap

比如下面直接通过AppInfo的totalCount比较实现排序

public static void main(String[] args) {
    AppInfo app1 = new AppInfo();
    app1.id = "id-1";
    app1.totalCount = 1;

    AppInfo app2 = new AppInfo();
    app2.id = "id-2";
    app2.totalCount = 2;

    AppInfo app3 = new AppInfo();
    app3.id = "id-3";
    app3.totalCount = 2;

    //java8的lambda创建Comparator
    Map<AppInfo, String> map = new TreeMap<>((left, right) -> left.totalCount - right.totalCount);
    map.put(app1, "value1");
    map.put(app2, "value2");
    map.put(app3, "value3");

    for (AppInfo app : map.keySet()) {
        System.out.println(String.format("[%s, %d] --> %s", app.id, app.totalCount, map.get(app)));
    }
}

static class AppInfo {
    public String id;
    public int totalCount;
}

结果为

[id-1, 1] --> value1

[id-2, 2] --> value3

关于Set和Map的注意事项

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