【发布时间】：2015-06-05 21:40:01
【问题描述】：
我正在尝试构建一个需要保存键和值的通用数据结构，同时跟踪放置键和值的索引，就像 arraylist 一样，复杂度为 O(log n)或更少。
我试图解决这个问题，并创建了各种具有整数的 TreeMap——索引和值，反之亦然，键也是如此。
为了更清楚，索引象征着用户的插入顺序。所以如果我有 3 个元素，那么它们的索引是 0 1 2，如果元素 0 被删除，那么我需要将 1 推到 0 并将 2 推到 1 并且将使用索引 2 添加一个新元素。
我的问题是当我删除一个键和它的值时，如果我想在正确的索引中插入下一个键和值，我必须确保所有旧的都设置回 1。我不知道如何做到这一点，而不是陷入 O(n) 复杂性。
我的目标是使用现有的数据结构并将它们混合以获得此结果，请查看我根据需要实现的方法。
我正在添加我的代码以供参考，任何帮助将不胜感激。
谢谢
汤姆
import java.util.Collection;
import java.util.HashMap;
import java.util.TreeMap;
import java.util.Map;
public class SuperStruct<K,V>
{
private Map<K,V> mInternalKeyToValueMap;//all the keys and their values
private Map<Integer,V> mIndexToValueMap;//index's for values according to the entrance order to
private Map<V,Integer> mValueToIndexMap;//values and their index's
private Map<Integer,K> mIndexToKeyMap;//index's and their keys according to entrance order
private Map<K,Integer> mKeyToIndexMap;//keys and their index's
private int mNextIndex;//index for the data structure according to the order data was received by user
public SuperStruct(){
mInternalKeyToValueMap = new TreeMap<K,V>();
mIndexToValueMap = new TreeMap<Integer,V>();
mValueToIndexMap = new TreeMap <V,Integer>();
mIndexToKeyMap = new TreeMap <Integer, K>();
mKeyToIndexMap = new TreeMap <K,Integer>();
}
public boolean containsKey(Object key) {
boolean containable = mInternalKeyToValueMap.containsKey(key);
return containable;
}
public boolean containsValue(Object value) {
boolean containable = mValueToIndexMap.containsKey(value);
return containable;
}
public V get(Object key) {
if(mInternalKeyToValueMap.containsKey(key)){
V value = mInternalKeyToValueMap.get(key);
return value;
}
return null;
}
public Collection<K> keySet() {
return mInternalKeyToValueMap.keySet();
}
/**
* This method is putting the key and the value in the main TreeMap "mInternalKeyToValueMap", while on the mean time updating 4 other TreeMaps
* with data regarding to the index in which data was received from the user.
* all in all this method runs in complexity of 6*(O(log n)) that sums down to O(log n) cause constants don't calculate over the whole
* Complexity calculation
* In case that a key already had a mapping to it and we overwrite the value we will run in complexity of 11*(O(log n)) which still sums down to O(log n)
* cause constants don't calculate over the whole
*/
public V put(K key, V value) {
if(mValueToIndexMap.containsKey(value))//preventing duplications of value
return value;
if(mInternalKeyToValueMap.containsKey(key)){//when a key already exist in system and we want to overwrite its value
int indexToDelete = mKeyToIndexMap.get(key);//we get the index of the value we over-write
V value1 = mIndexToValueMap.get(indexToDelete);//using this index we get the value
mValueToIndexMap.remove(value1);//we remove the value and its index
mIndexToValueMap.remove(indexToDelete);//we remove the index and its value
}
mInternalKeyToValueMap.put(key, value);//putting the new value for the key in the main TreeMap
mValueToIndexMap.put(value, mNextIndex);//populating the TreeMap of values and their index's - the order we received them from the user
mIndexToValueMap.put(mNextIndex, value);//This TreeMap holds the index's for each value according to the order of insertion by user
mIndexToKeyMap.put(mNextIndex, key);//This TreeMap holds the index's for each key according to the order of insertion by user
mKeyToIndexMap.put(key,mNextIndex);//populating the TreeMap of keys and their index's - the order we received them from the user
++mNextIndex;//advancing the index which mark the insertion order of arguments to structure
return null;
}
public V remove(Object key) {
if(mInternalKeyToValueMap.containsKey(key)==true && (mInternalKeyToValueMap.get(key)!=null))
{
V value = mInternalKeyToValueMap.get(key);
mInternalKeyToValueMap.remove(key);//removing map for the value
int mIndexToRemoveValue = mValueToIndexMap.get(value);//getting the right index to remove the value
mIndexToValueMap.remove(mIndexToRemoveValue);//vacating the value for this index
mIndexToKeyMap.remove(mIndexToRemoveValue);//vacating the key for this index
mKeyToIndexMap.remove(key);//removing a key and index in the keyToIndex Map
mValueToIndexMap.remove(value);//removing a key and index in the ValueToIndex Map
return value;
}
return null;
}
public Collection<V> values() {
return mInternalKeyToValueMap.values();
}
public Collection<V> getStrcutureSorted(){
return mValueToIndexMap.keySet();
}
public V getValueByIndex(int index){//return the index in which the value sits in, if not present null
V value = mIndexToValueMap.get(index);
return value;
}
public Integer getIndexByKey(K key){
Integer returnable = mKeyToIndexMap.get(key);
if(returnable == null)
return -1;
return returnable;
}
public K getKeyByIndex(int index){
return mIndexToKeyMap.get(index);
}
}
【问题讨论】：
标签： arraylist indexing data-structures time-complexity treemap