【发布时间】:2010-11-22 02:00:06
【问题描述】:
【问题讨论】:
-
@Click: "...对于所有具有特定值的键..."。您是指给定 Set 中的所有键,还是满足给定谓词的所有键?
标签: java collections treemap
【问题讨论】:
标签: java collections treemap
//create TreeMap instance
TreeMap treeMap = new TreeMap();
//add key value pairs to TreeMap
treeMap.put("1","One");
treeMap.put("2","Two");
treeMap.put("3","Three");
/*
get Collection of values contained in TreeMap using
Collection values()
*/
Collection c = treeMap.values();
//obtain an Iterator for Collection
Iterator itr = c.iterator();
//iterate through TreeMap values iterator
while(itr.hasNext())
System.out.println(itr.next());
或:
for (Map.Entry<K,V> entry : treeMap.entrySet()) {
V value = entry.getValue();
K key = entry.getKey();
}
或:
// Use iterator to display the keys and associated values
System.out.println("Map Values Before: ");
Set keys = map.keySet();
for (Iterator i = keys.iterator(); i.hasNext();) {
Integer key = (Integer) i.next();
String value = (String) map.get(key);
System.out.println(key + " = " + value);
}
【讨论】:
只是指出迭代任何地图的通用方法:
private <K, V> void iterateOverMap(Map<K, V> map) {
for (Map.Entry<K, V> entry : map.entrySet()) {
System.out.println("key ->" + entry.getKey() + ", value->" + entry.getValue());
}
}
【讨论】:
使用Google Collections,假设 K 是您的密钥类型:
Maps.filterKeys(treeMap, new Predicate<K>() {
@Override
public boolean apply(K key) {
return false; //return true here if you need the entry to be in your new map
}});
如果您也需要该值,也可以使用 filterEntries。
【讨论】:
假设类型 TreeMap
for(Map.Entry<String,Integer> entry : treeMap.entrySet()) {
String key = entry.getKey();
Integer value = entry.getValue();
System.out.println(key + " => " + value);
}
(key 和 Value 类型当然可以是任何类)
【讨论】:
treeMap 给你的 TreeMap 起初让我失望。