java - 将对象作为参数传递并在方法内对其进行修改

Question

假设我有一个 Map<String, String>我想删除 value 包含的所有条目 foo .就优化/内存/等而言，最好的方法是什么？四syso下面打印相同的结果，也就是说 {n2=bar} .

public static void main(String[] args) {

    Map<String, String> in = new HashMap<String, String>();
    in.put("n1", "foo");
    in.put("n2", "bar");
    in.put("n3", "foobar");

    // 1- create a new object with the returned Map
    Map<String, String> in1 = new HashMap<String, String>(in);
    Map<String, String> out1 = methodThatReturns(in1);
    System.out.println(out1);

    // 2- overwrite the initial Map with the returned one 
    Map<String, String> in2 = new HashMap<String, String>(in);
    in2 = methodThatReturns(in2);
    System.out.println(in2);

    // 3- use the clear/putAll methods
    Map<String, String> in3 = new HashMap<String, String>(in);
    methodThatClearsAndReadds(in3);
    System.out.println(in3);

    // 4- use an iterator to remove elements
    Map<String, String> in4 = new HashMap<String, String>(in);
    methodThatRemoves(in4);
    System.out.println(in4);

}

public static Map<String, String> methodThatReturns(Map<String, String> in) {
    Map<String, String> out = new HashMap<String, String>();
    for(Entry<String, String> entry : in.entrySet()) {
        if(!entry.getValue().contains("foo")) {
            out.put(entry.getKey(), entry.getValue());
        }
    }
    return out;
}

public static void methodThatClearsAndReadds(Map<String, String> in) {
    Map<String, String> out = new HashMap<String, String>();
    for(Entry<String, String> entry : in.entrySet()) {
        if(!entry.getValue().contains("foo")) {
            out.put(entry.getKey(), entry.getValue());
        }
    }
    in.clear();
    in.putAll(out);
}

public static void methodThatRemoves(Map<String, String> in) {
    for(Iterator<Entry<String, String>> it = in.entrySet().iterator(); it.hasNext();) {
        if(it.next().getValue().contains("foo")) {
            it.remove();
        }
    }
}

Answer 1

最好的方法是 methodThatRemoves 因为:

1. 在内存消耗方面:它不会创建新 map ，因此不会增加内存开销。
2. 在 CPU 使用方面:迭代器在调用 next 或删除当前元素时具有 O(1) 复杂度。