Map迭代问题

Question

我想迭代Map,其中值是另一个Map.我有这个方法:

 public void test(){
   Map<Integer, Double> map1 = new HashMap<Integer, Double>();
   Map<Integer, Map<Integer, Double>> map2 = new HashMap<Integer, Map<Integer, Double>>();
   map1.put(1,1.1);
   map1.put(2,2.2);
   map1.put(3,3.3);
   map2.put(1,map1);
   map1.clear();
   map1.put(4,4.4);
   map1.put(5,5.5);
   map1.put(6,6.6);
   map2.put(2,map1);

   for (Map.Entry<Integer, Map<Integer, Double>> entry : map2.entrySet()){
       Log.d("test", "map2 key " + entry.getKey());
       Map<Integer, Double> tempMap = new HashMap<Integer, Double>(entry.getValue());
       for (Map.Entry<Integer, Double> entry1 : tempMap.entrySet()){
           Log.d("map", "map1 key " + entry1.getKey() + "map1 value " + entry1.getValue());
       }
   }

我在日志中看到这个:

map2 key 1
map1 key 4 map1 value 4.4
map1 key 5 map1 value 5.5
map1 key 6 map1 value 6.6
map2 key 2
map1 key 4 map1 value 4.4
map1 key 5 map1 value 5.5
map1 key 6 map1 value 6.6

有什么问题？

Answer 1

我怀疑你想要钥匙1,2和3 map1.

问题是您重复使用了相同的地图map1,并将其插入map2两次.因此,所有操作的累积结果map1是地图中只有4,5和6.视觉:

   /1 ====v
map2    map1 -> (4 => 4.4), (5 => 5.5), (6 => 6.6)
   \2 ====^

要获得两个不同的地图,请map1在第一次插入后清除,创建新地图.更换

map1.clear();

同

map1 = new HashMap<Integer, Double>();

这将产生这样的:

   /1====> (1 => 1.1), (2 => 2.2), (3 => 3.3)
map2    
   \2====> (4 => 4.4), (5 => 5.5), (6 => 6.6)
            ^
map1  ------/