Java中的Map如果在遍历过程中要删除元素，除非通过迭代器自己的remove()方法，否则就会导致抛出ConcurrentModificationException异常。JDK文档中是这么描述的：

The iterators returned by all of this class’s “collection view methods” are fail-fast: if the map is structurally modified at any time after the iterator is created, in any way except through the iterator’s own remove method, the iterator will throw a ConcurrentModificationException. Thus, in the face of concurrent modification, the iterator fails quickly and cleanly, rather than risking arbitrary, non-deterministic behavior at an undetermined time in the future.

这么做的原因是为了保证迭代器能够尽快感知到Map的“结构性修改“，从而避免不同视图下不一致现象。

下面代码展示了遍历Map时删除元素的正确方式和错误方式。

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Set;

/**
 * Created by lh on 15-1-22.
 */
public class TestMapRemove {
    public static void main(String[] args){
        new TestMapRemove().removeByIterator();
//        new TestMapRemove().removeBymap();
    }
    public void removeByIterator(){//正确的删除方式
        HashMap<Integer, String> map = new HashMap<Integer, String>();
        map.put(1, "one");
        map.put(2, "two");
        map.put(3, "three");
        System.out.println(map);
        Iterator<Map.Entry<Integer, String>> it = map.entrySet().iterator();
        while(it.hasNext()){
            Map.Entry<Integer, String> entry = it.next();
            if(entry.getKey() == 2)
                it.remove();//使用迭代器的remove()方法删除元素
        }
        System.out.println(map);
    }
    public void removeBymap(){//错误的删除方式
        HashMap<Integer, String> map = new HashMap<Integer, String>();
        map.put(1, "one");
        map.put(2, "two");
        map.put(3, "three");
        System.out.println(map);
        Set<Map.Entry<Integer, String>> entries = map.entrySet();
        for(Map.Entry<Integer, String> entry : entries){
            if(entry.getKey() == 2){
                map.remove(entry.getKey());//ConcurrentModificationException
            }
        }
        System.out.println(map);
    }
}