关于map的一个小问题

[gcw634]

List listA = new ArrayList();
		List listB = new ArrayList();

		Object[] objectc = new Object[] { "aa", "123", "542" };
		Object[] objecte = new Object[] { "ww", "524", "865" };
		Object[] objecth = new Object[] { "qq", "458", "632" };
		Object[] objectp = new Object[] { "gg", "758", "632" };

		listA.add(objectc);
		listA.add(objecte);
		listA.add(objecth);
		listA.add(objectp);

		Object[] objectq = new Object[] { "gg", "632", "751" };
		Object[] objectb = new Object[] { "aa", "123", "123" };
		Object[] objecty = new Object[] { "gg", "952", "156" };

		listB.add(objectq);
		listB.add(objectb);
		listB.add(objecty);

		Hashtable<String, LinkedHashSet<Object>> ht = new Hashtable<String, LinkedHashSet<Object>>();
		for (int i = 0; i < listA.size(); i++) {
			Object[] ss = (Object[]) (listA.get(i));
			LinkedHashSet<Object> set = new LinkedHashSet<Object>();
			set.addAll(Arrays.asList(ss));
			ht.put((String)ss[0], set);
		}

		for (int i = 0; i < listB.size(); i++) {
			Object[] ss = (Object[]) (listB.get(i));
			LinkedHashSet<Object> set = new LinkedHashSet<Object>();
			set.addAll(Arrays.asList(ss));
			if (ht.get((String)ss[0]) != null) {
				LinkedHashSet<Object> value = ht.get(ss[0]);
				value.addAll(set);
				ht.put((String)ss[0], value);
			} else {
				ht.put((String)ss[0], set);
			}
		}

		Iterator it = ht.keySet().iterator();
		while (it.hasNext()) {
			String key = (String) (it.next());
			LinkedHashSet value = ht.get(key);
			Iterator it1 = value.iterator();
			while (it1.hasNext()) {
				System.out.print(it1.next() + " ");
			}
			System.out.println();
		}

结果是这样：

qq 458 632 
gg 758 632 751 952 156 
ww 524 865 
aa 123 542  

我要的结果是：

qq 458 632 
gg 758 632 632 751 952 156 
ww 524 865 
aa 123 542 123 123 

为什么我listA和listB中的重复的值都没有了，我这个值是value啊，并不是key

[focusforce]

LinkedHashSet造成的，你用ArrayList就是你要的结果。

[qianzhimeiying]

...这好象是我刚才写的那个...

value我定义成了set,所以addAll的时候重复的部分就去掉了,你不希望去掉重复的,就换成ArrayList

[gcw633]

[Quote=引用 2 楼 qianzhimeiying 的回复:]
...这好象是我刚才写的那个...

value我定义成了set,所以addAll的时候重复的部分就去掉了,你不希望去掉重复的,就换成ArrayList
[/Quote]
是你写的，谢谢啊！！没有分了，所以又开了个小号来求教，第一次使用哈希表，很多都不明白呢

[healer_kx]

这个对了。。。

是因为LinkedHashSet造成的，他存放的内容不可以重复。

import java.lang.reflect.Constructor;
import java.lang.reflect.InvocationTargetException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Hashtable;
import java.util.Iterator;
import java.util.LinkedHashSet;
import java.util.List;

public class Cast {

	public static void main(String[] args) {
		List listA = new ArrayList();
		List listB = new ArrayList();
		Object[] objectc = new Object[] { "aa", "123", "542" };
		Object[] objecte = new Object[] { "ww", "524", "865" };
		Object[] objecth = new Object[] { "qq", "458", "632" };
		Object[] objectp = new Object[] { "gg", "758", "632" };
		listA.add(objectc);
		listA.add(objecte);
		listA.add(objecth);
		listA.add(objectp);
		Object[] objectq = new Object[] { "gg", "632", "751" };
		Object[] objectb = new Object[] { "aa", "123", "123" };
		Object[] objecty = new Object[] { "gg", "952", "156" };
		listB.add(objectq);
		listB.add(objectb);
		listB.add(objecty);
		Hashtable<String, ArrayList<Object>> ht = new Hashtable<String, ArrayList<Object>>();
		for (int i = 0; i < listA.size(); i++) {
			Object[] ss = (Object[]) (listA.get(i));
			ArrayList<Object> set = new ArrayList<Object>();
			set.addAll(Arrays.asList(ss));
			ht.put((String) ss[0], set);
		}
		for (int i = 0; i < listB.size(); i++) {
			Object[] ss = (Object[]) (listB.get(i));
			ArrayList<Object> set = new ArrayList<Object>();
			set.addAll(Arrays.asList(Arrays.copyOfRange(ss, 1, ss.length)));
			if (ht.get((String) ss[0]) != null) {
				
				ArrayList<Object> value = ht.get(ss[0]);
				value.addAll(set);
				ht.put((String) ss[0], value);
			} else {
				ht.put((String) ss[0], set);
			}
		}
		Iterator it = ht.keySet().iterator();
		while (it.hasNext()) {
			String key = (String) (it.next());
			ArrayList value = ht.get(key);
			Iterator it1 = value.iterator();
			while (it1.hasNext()) {
				System.out.print(it1.next() + " ");
			}
			System.out.println();
		}

	}
}