因为containsKey() 只是一个 get() 丢弃了检索到的值,所以它是 O(1)(再次假设哈希函数正常工作)。 原文由 Michael Borgwardt 发布,翻译遵循 CC BY-SA 3.0 许可协议 有用 回复 社区维基1 发布于 2022-11-24 通常是 O(1),但如果我们使用了一个错误的 hashCode 函数,我们需要将多个元素添加到一个
问HashMap.containsKey()在java中的时间复杂度是多少？EN1，当 i = 0 时，内循环执行 n 次运算，...
1. containsKey() 判断HashMap是否包含key 2. containsValue() 判断HashMap是否包含“值为value”的元素 3. get() 获取key对应的value 4. put() 让HashMap对象可以通过put()将“key-value”添加到HashMap中 5. remove() 删除“键为key”元素 遍历方式 1. 在for-each循环中使用entries来遍历 1 Map<Integer...
API时间复杂度: add(E e)是在尾巴上加元素,虽然 ArrayList 可能会有扩容的情况出现,但是均摊复杂度(amortized time complexity)还是 O(1) 的。 add(int index, E e)是在特定的位置上加元素,LinkedList 需要先找到这个位置,再加上这个元素,虽然单纯的「加」这个动作是 O(1) 的,但是要找到这个位置还是 O(n...
{ return containsKey(o); } public final boolean remove(Object key) { return removeNode(hash(key), key, null, false, true) != null; } public final Spliterator<K> spliterator() { return new KeySpliterator<>(HashMap.this, 0, -1, 0, 0); } public Object[] toArray() { return keys...
importjava.util.HashMap;publicclassHashMapExample{publicstaticvoidmain(String[]args){HashMap<String,Integer>map=newHashMap<>();map.put("Alice",25);map.put("Bob",30);map.put("Charlie",35);intage=map.get("Bob");System.out.println("Bob's age is: "+age);booleanexists=map.containsKey(...
To check if a key is present in the map, we can use thecontainsKey()method: Or, to check if a value is present in the map, we can use thecontainsValue()method: Both method calls will returntruein our example. Though they look very similar, there is an important difference in perform...
</li> * <li> Runtime complexity of {@code O(log(n)}). </li> * </ul> * @param key Integer key for the value * @param value Double value of the key */ public void put(int key, double value) { if (!_keyToIndexMap.containsKey(key)) { _values.add(value); int last = _...
The combination ofcontainsKeyandputmethods is another way to update the value of a key inHashMap.This option checks if the map already contains a key. In such a case, we can update the value using theputmethod.Otherwise, we can either add an entry to the map or do nothing. ...
which is normally the order in which keys were inserted into the map (insertion-order). Note that insertion order is not affected if a key is re-inserted into the map. (A key k is reinserted into a map m if m.put(k, v) is invoked when m.containsKey(k) would return true immediat...