问题描述

我有一个如下的哈希图:

I have a hashmap as below:

1->x

2->是

3->x

4->z

现在我想知道值为 x 的所有键(ans: [1,3] ).最好的方法是什么?

Now i want to know all keys whose value is x (ans: [1,3] ). what is best way to do?

蛮力方法是遍历 map 并将所有键存储在值为 x 的数组中.

Brute force way is to just iterate over map and store all keys in array whose value is x.

有什么有效的方法吗?

谢谢

推荐答案

你可以使用 MultiMap 轻松获取所有这些重复值.

You can use a MultiMap to easily get all those duplicate values.

Map<Integer, String> map = new HashMap<Integer, String>();
map.put(1, "x");
map.put(2, "y");
map.put(2, "z");
map.put(3, "x");
map.put(4, "y");
map.put(5, "z");
map.put(6, "x");
map.put(7, "y");

System.out.println("Original map: " + map);

Multimap<String, Integer> multiMap = HashMultimap.create();
for (Entry<Integer, String> entry : map.entrySet()) {
  multiMap.put(entry.getValue(), entry.getKey());
}
System.out.println();

for (Entry<String, Collection<Integer>> entry : multiMap.asMap().entrySet()) {
  System.out.println("Original value: " + entry.getKey() + " was mapped to keys: "
      + entry.getValue());
}

打印出来:

Original map: {1=x, 2=z, 3=x, 4=y, 5=z, 6=x, 7=y}

Original value: z was mapped to keys: [2, 5]
Original value: y was mapped to keys: [4, 7]
Original value: x was mapped to keys: [1, 3, 6]

根据@noahz 的建议，forMap 和 invertFrom 需要的行数更少，但阅读起来可能更复杂:

Per @noahz's suggestion, forMap and invertFrom takes fewer lines, but is arguably more complex to read:

HashMultimap<String, Integer> multiMap =
    Multimaps.invertFrom(Multimaps.forMap(map), 
        HashMultimap.<String, Integer> create());

代替:

Multimap<String, Integer> multiMap = HashMultimap.create();
for (Entry<Integer, String> entry : map.entrySet()) {
  multiMap.put(entry.getValue(), entry.getKey());
}

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