这个问题之前在 Stack Exchange 上被问过,但没有得到答案。
之前的问题链接: 二叉树结构实现二叉堆
如何在二叉树中实现堆。要实现堆,需要知道最后一个已填充节点和第一个未占用节点。这可以通过树的层序遍历来完成,但是如果这样做,则找到第一个未占用节点的时间复杂度将为 O(n)。所以,如何在二叉树中以 O(logn) 的时间复杂度实现堆呢?
谢谢 Shekhar
这个问题之前在 Stack Exchange 上被问过,但没有得到答案。
之前的问题链接: 二叉树结构实现二叉堆
如何在二叉树中实现堆。要实现堆,需要知道最后一个已填充节点和第一个未占用节点。这可以通过树的层序遍历来完成,但是如果这样做,则找到第一个未占用节点的时间复杂度将为 O(n)。所以,如何在二叉树中以 O(logn) 的时间复杂度实现堆呢?
谢谢 Shekhar
你不会将堆实现在二叉树中,因为堆是一个二叉树。堆维护以下顺序属性 - 给定节点V,其父节点大于或等于V。此外,堆是完整的二叉树。我在大学修过ADS课程,所以稍后我会在答案中提供我的Java堆实现。只列出您获得的主要方法复杂度:
这是我的Heap.java
文件:
public class Heap<E extends Comparable<E>> {
private Object S[];
private int last;
private int capacity;
public Heap() {
S = new Object[11];
last = 0;
capacity = 7;
}
public Heap(int cap) {
S = new Object[cap + 1];
last = 0;
capacity = cap;
}
public int size() {
return last;
}
//
// returns the number of elements in the heap
//
public boolean isEmpty() {
return size() == 0;
}
//
// is the heap empty?
//
public E min() throws HeapException {
if (isEmpty())
throw new HeapException("The heap is empty.");
else
return (E) S[1];
}
//
// returns element with smallest key, without removal
//
private int compare(Object x, Object y) {
return ((E) x).compareTo((E) y);
}
public void insert(E e) throws HeapException {
if (size() == capacity)
throw new HeapException("Heap overflow.");
else{
last++;
S[last] = e;
upHeapBubble();
}
}
// inserts e into the heap
// throws exception if heap overflow
//
public E removeMin() throws HeapException {
if (isEmpty())
throw new HeapException("Heap is empty.");
else {
E min = min();
S[1] = S[last];
last--;
downHeapBubble();
return min;
}
}
//
// removes and returns smallest element of the heap
// throws exception is heap is empty
//
/**
* downHeapBubble() method is used after the removeMin() method to reorder the elements
* in order to preserve the Heap properties
*/
private void downHeapBubble(){
int index = 1;
while (true){
int child = index*2;
if (child > size())
break;
if (child + 1 <= size()){
//if there are two children -> take the smalles or
//if they are equal take the left one
child = findMin(child, child + 1);
}
if (compare(S[index],S[child]) <= 0 )
break;
swap(index,child);
index = child;
}
}
/**
* upHeapBubble() method is used after the insert(E e) method to reorder the elements
* in order to preserve the Heap properties
*/
private void upHeapBubble(){
int index = size();
while (index > 1){
int parent = index / 2;
if (compare(S[index], S[parent]) >= 0)
//break if the parent is greater or equal to the current element
break;
swap(index,parent);
index = parent;
}
}
/**
* Swaps two integers i and j
* @param i
* @param j
*/
private void swap(int i, int j) {
Object temp = S[i];
S[i] = S[j];
S[j] = temp;
}
/**
* the method is used in the downHeapBubble() method
* @param leftChild
* @param rightChild
* @return min of left and right child, if they are equal return the left
*/
private int findMin(int leftChild, int rightChild) {
if (compare(S[leftChild], S[rightChild]) <= 0)
return leftChild;
else
return rightChild;
}
public String toString() {
String s = "[";
for (int i = 1; i <= size(); i++) {
s += S[i];
if (i != last)
s += ",";
}
return s + "]";
}
//
// outputs the entries in S in the order S[1] to S[last]
// in same style as used in ArrayQueue
//
}
HeapException.java:
public class HeapException extends RuntimeException {
public HeapException(){};
public HeapException(String msg){super(msg);}
}
downHeapBubble()
和upHeapBubble()
方法,它们使性能达到O(logn)。我很快会添加关于它们的好的解释。upHeapBubble()
方法。因此,当您插入最后一个位置时,需要调用upHeapBubble()
方法,如下所示:last++;
S[last] = e;
upHeapBubble();
然后将最后一个元素与其父元素进行比较,如果父元素更大,则交换:这将最多执行logn次,其中n是节点数。因此,这里就有了logn的性能。
对于删除部分-您只能删除最小值-最高节点。因此,当您将其删除时,必须将其与最后一个节点交换-但然后您必须维护堆属性并执行downHeapBubble()
。如果节点大于其子节点,请与最小的子节点交换,依此类推,直到没有子节点或没有更小的子节点为止。这可以最多执行logn次,因此这里就有了logn的性能。您可以通过查看二叉树图片here来自行解释为什么可以最多执行此操作logn次。
使用树来实现堆
我自己回答一个关于堆的问题,它需要O(log n)时间复杂度,但限制条件是要保留父节点指针。如果我们不保留父节点指针,那么需要大约O(n)时间复杂度。我发布了这个问题是为了寻求O(log n)的解决方案。
以下是计算下一个未被占用叶子节点的步骤(我们拥有父节点的指针):
x = last inserted node. We save this after every insertion.
y = tmp node
z = next unoccupied node (next insertion)
if x is left child
z = x -> parent -> rightchild (problem solved.. that was easy)
else if x is right child
go to x's parent, until parent becomes left child. Let this node be y
(subtree rooted at y's sibling will contain the next unoccupied node)
z = y -> parent -> right -> go left until null
availableLeaf(node) {
if( node.left is Empty || node.right is Empty )
return node ;
else
if( node.left.count < node.right.count )
return availableLeaf(node.left)
else
return availableLeaf(node.right)
}
这种策略还可以平衡每个子树两侧节点的数量,对性能有益(尽管极其微小)。
这是O(log n)级别的。在插入时跟踪计数需要一直回到根节点,但这不会改变此操作的O(log n)级别特性。删除也是相似的情况。
其他操作都与通常的操作方式相同,并保持其性能特征。
你需要详细信息还是想自己解决?
如果您想使用仅具有左右指针而没有其他信息的链式二叉树,则建议您发起至少100,000分的悬赏。我并不是说这是不可能的(因为我没有数学证明),但我知道这已经有好几十年了。
import java.util.ArrayList;
import java.util.List;
/**
* @author Harish R
*/
public class HeapPractise<T extends Comparable<T>> {
private List<T> heapList;
public List<T> getHeapList() {
return heapList;
}
public void setHeapList(List<T> heapList) {
this.heapList = heapList;
}
private int heapSize;
public HeapPractise() {
this.heapList = new ArrayList<>();
this.heapSize = heapList.size();
}
public void insert(T item) {
if (heapList.size() == 0) {
heapList.add(item);
} else {
siftUp(item);
}
}
private void siftUp(T item) {
heapList.add(item);
heapSize = heapList.size();
int currentIndex = heapSize - 1;
while (currentIndex > 0) {
int parentIndex = (int) Math.floor((currentIndex - 1) / 2);
T parentItem = heapList.get(parentIndex);
if (parentItem != null) {
if (item.compareTo(parentItem) > 0) {
heapList.set(parentIndex, item);
heapList.set(currentIndex, parentItem);
currentIndex = parentIndex;
continue;
}
}
break;
}
}
public T delete() {
if (heapList.size() == 0) {
return null;
}
if (heapList.size() == 1) {
T item = heapList.get(0);
heapList.remove(0);
return item;
}
return siftDown();
}
private T siftDown() {
T item = heapList.get(0);
T lastItem = heapList.get(heapList.size() - 1);
heapList.remove(heapList.size() - 1);
heapList.set(0, lastItem);
heapSize = heapList.size();
int currentIndex = 0;
while (currentIndex < heapSize) {
int leftIndex = (2 * currentIndex) + 1;
int rightIndex = (2 * currentIndex) + 2;
T leftItem = null;
T rightItem = null;
int currentLargestItemIndex = -1;
if (leftIndex <= heapSize - 1) {
leftItem = heapList.get(leftIndex);
}
if (rightIndex <= heapSize - 1) {
rightItem = heapList.get(rightIndex);
}
T currentLargestItem = null;
if (leftItem != null && rightItem != null) {
if (leftItem.compareTo(rightItem) >= 0) {
currentLargestItem = leftItem;
currentLargestItemIndex = leftIndex;
} else {
currentLargestItem = rightItem;
currentLargestItemIndex = rightIndex;
}
} else if (leftItem != null && rightItem == null) {
currentLargestItem = leftItem;
currentLargestItemIndex = leftIndex;
}
if (currentLargestItem != null) {
if (lastItem.compareTo(currentLargestItem) >= 0) {
break;
} else {
heapList.set(currentLargestItemIndex, lastItem);
heapList.set(currentIndex, currentLargestItem);
currentIndex = currentLargestItemIndex;
continue;
}
} else {
break;
}
}
return item;
}
public static void main(String[] args) {
HeapPractise<Integer> heap = new HeapPractise<>();
for (int i = 0; i < 32; i++) {
heap.insert(i);
}
System.out.println(heap.getHeapList());
List<Node<Integer>> nodeArray = new ArrayList<>(heap.getHeapList()
.size());
for (int i = 0; i < heap.getHeapList().size(); i++) {
Integer heapElement = heap.getHeapList().get(i);
Node<Integer> node = new Node<Integer>(heapElement);
nodeArray.add(node);
}
for (int i = 0; i < nodeArray.size(); i++) {
int leftNodeIndex = (2 * i) + 1;
int rightNodeIndex = (2 * i) + 2;
Node<Integer> node = nodeArray.get(i);
if (leftNodeIndex <= heap.getHeapList().size() - 1) {
Node<Integer> leftNode = nodeArray.get(leftNodeIndex);
node.left = leftNode;
}
if (rightNodeIndex <= heap.getHeapList().size() - 1) {
Node<Integer> rightNode = nodeArray.get(rightNodeIndex);
node.right = rightNode;
}
}
BTreePrinter.printNode(nodeArray.get(0));
System.out.println(heap.delete());
nodeArray = new ArrayList<>(heap.getHeapList().size());
for (int i = 0; i < heap.getHeapList().size(); i++) {
Integer heapElement = heap.getHeapList().get(i);
Node<Integer> node = new Node<Integer>(heapElement);
nodeArray.add(node);
}
for (int i = 0; i < nodeArray.size(); i++) {
int leftNodeIndex = (2 * i) + 1;
int rightNodeIndex = (2 * i) + 2;
Node<Integer> node = nodeArray.get(i);
if (leftNodeIndex <= heap.getHeapList().size() - 1) {
Node<Integer> leftNode = nodeArray.get(leftNodeIndex);
node.left = leftNode;
}
if (rightNodeIndex <= heap.getHeapList().size() - 1) {
Node<Integer> rightNode = nodeArray.get(rightNodeIndex);
node.right = rightNode;
}
}
BTreePrinter.printNode(nodeArray.get(0));
}
}
public class Node<T extends Comparable<?>> {
Node<T> left, right;
T data;
public Node(T data) {
this.data = data;
}
}
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
class BTreePrinter {
public static <T extends Comparable<?>> void printNode(Node<T> root) {
int maxLevel = BTreePrinter.maxLevel(root);
printNodeInternal(Collections.singletonList(root), 1, maxLevel);
}
private static <T extends Comparable<?>> void printNodeInternal(
List<Node<T>> nodes, int level, int maxLevel) {
if (nodes.isEmpty() || BTreePrinter.isAllElementsNull(nodes))
return;
int floor = maxLevel - level;
int endgeLines = (int) Math.pow(2, (Math.max(floor - 1, 0)));
int firstSpaces = (int) Math.pow(2, (floor)) - 1;
int betweenSpaces = (int) Math.pow(2, (floor + 1)) - 1;
BTreePrinter.printWhitespaces(firstSpaces);
List<Node<T>> newNodes = new ArrayList<Node<T>>();
for (Node<T> node : nodes) {
if (node != null) {
String nodeData = String.valueOf(node.data);
if (nodeData != null) {
if (nodeData.length() == 1) {
nodeData = "0" + nodeData;
}
}
System.out.print(nodeData);
newNodes.add(node.left);
newNodes.add(node.right);
} else {
newNodes.add(null);
newNodes.add(null);
System.out.print(" ");
}
BTreePrinter.printWhitespaces(betweenSpaces);
}
System.out.println("");
for (int i = 1; i <= endgeLines; i++) {
for (int j = 0; j < nodes.size(); j++) {
BTreePrinter.printWhitespaces(firstSpaces - i);
if (nodes.get(j) == null) {
BTreePrinter.printWhitespaces(endgeLines + endgeLines + i
+ 1);
continue;
}
if (nodes.get(j).left != null)
System.out.print("//");
else
BTreePrinter.printWhitespaces(1);
BTreePrinter.printWhitespaces(i + i - 1);
if (nodes.get(j).right != null)
System.out.print("\\\\");
else
BTreePrinter.printWhitespaces(1);
BTreePrinter.printWhitespaces(endgeLines + endgeLines - i);
}
System.out.println("");
}
printNodeInternal(newNodes, level + 1, maxLevel);
}
private static void printWhitespaces(int count) {
for (int i = 0; i < 2 * count; i++)
System.out.print(" ");
}
private static <T extends Comparable<?>> int maxLevel(Node<T> node) {
if (node == null)
return 0;
return Math.max(BTreePrinter.maxLevel(node.left),
BTreePrinter.maxLevel(node.right)) + 1;
}
private static <T> boolean isAllElementsNull(List<T> list) {
for (Object object : list) {
if (object != null)
return false;
}
return true;
}
}
我的堆实现
public class Heap <T extends Comparable<T>> {
private T[] arr;
private int size;
public Heap(T[] baseArr) {
this.arr = baseArr;
size = arr.length - 1;
}
public void minHeapify(int i, int n) {
int l = 2 * i + 1;
int r = 2 * i + 2;
int smallest = i;
if (l <= n && arr[l].compareTo(arr[smallest]) < 0) {
smallest = l;
}
if (r <= n && arr[r].compareTo(arr[smallest]) < 0) {
smallest = r;
}
if (smallest != i) {
T temp = arr[i];
arr[i] = arr[smallest];
arr[smallest] = temp;
minHeapify(smallest, n);
}
}
public void buildMinHeap() {
for (int i = size / 2; i >= 0; i--) {
minHeapify(i, size);
}
}
public void heapSortAscending() {
buildMinHeap();
int n = size;
for (int i = n; i >= 1; i--) {
T temp = arr[0];
arr[0] = arr[i];
arr[i] = temp;
n--;
minHeapify(0, n);
}
}
}
二叉树可以用数组表示:
import java.util.Arrays;
public class MyHeap {
private Object[] heap;
private int capacity;
private int size;
public MyHeap() {
capacity = 8;
heap = new Object[capacity];
size = 0;
}
private void increaseCapacity() {
capacity *= 2;
heap = Arrays.copyOf(heap, capacity);
}
public int getSize() {
return size;
}
public boolean isEmpty() {
return size == 0;
}
public Object top() {
return size > 0 ? heap[0] : null;
}
@SuppressWarnings("unchecked")
public Object remove() {
if (size == 0) {
return null;
}
size--;
Object res = heap[0];
Object te = heap[size];
int curr = 0, son = 1;
while (son < size) {
if (son + 1 < size
&& ((Comparable<Object>) heap[son + 1])
.compareTo(heap[son]) < 0) {
son++;
}
if (((Comparable<Object>) te).compareTo(heap[son]) <= 0) {
break;
}
heap[curr] = heap[son];
curr = son;
son = 2 * curr + 1;
}
heap[curr] = te;
return res;
}
@SuppressWarnings("unchecked")
public void insert(Object e) {
if (size == capacity) { // auto scaling
increaseCapacity();
}
int curr = size;
int parent;
heap[size] = e;
size++;
while (curr > 0) {
parent = (curr - 1) / 2;
if (((Comparable<Object>) heap[parent]).compareTo(e) <= 0) {
break;
}
heap[curr] = heap[parent];
curr = parent;
}
heap[curr] = e;
}
}
使用方法:
MyHeap heap = new MyHeap(); // it is a min heap
heap.insert(18);
heap.insert(26);
heap.insert(35);
System.out.println("size is " + heap.getSize() + ", top is " + heap.top());
heap.insert(36);
heap.insert(30);
heap.insert(10);
while(!heap.isEmpty()) {
System.out.println(heap.remove());
}
这里是 - 使用二叉树实现堆的代码
public class PriorityQ<K extends Comparable<K>> {
private class TreeNode<T extends Comparable<T>> {
T val;
TreeNode<T> left, right, parent;
public String toString() {
return this.val.toString();
}
TreeNode(T v) {
this.val = v;
left = null;
right = null;
}
public TreeNode<T> insert(T val, int position) {
TreeNode<T> parent = findNode(position/2);
TreeNode<T> node = new TreeNode<T>(val);
if(position % 2 == 0) {
parent.left = node;
} else {
parent.right = node;
}
node.parent = parent;
heapify(node);
return node;
}
private void heapify(TreeNode<T> node) {
while(node.parent != null && (node.parent.val.compareTo(node.val) < 0)) {
T temp = node.val;
node.val = node.parent.val;
node.parent.val = temp;
node = node.parent;
}
}
private TreeNode<T> findNode(int pos) {
TreeNode<T> node = this;
int reversed = 1;
while(pos > 0) {
reversed <<= 1;
reversed |= (pos&1);
pos >>= 1;
}
reversed >>= 1;
while(reversed > 1) {
if((reversed & 1) == 0) {
node = node.left;
} else {
node = node.right;
}
reversed >>= 1;
}
return node;
}
public TreeNode<T> remove(int pos) {
if(pos <= 1) {
return null;
}
TreeNode<T> last = findNode(pos);
if(last.parent.right == last) {
last.parent.right = null;
} else {
last.parent.left = null;
}
this.val = last.val;
bubbleDown();
return null;
}
public void bubbleDown() {
TreeNode<T> node = this;
do {
TreeNode<T> left = node.left;
TreeNode<T> right = node.right;
if(left != null && right != null) {
T max = left.val.compareTo(right.val) > 0 ? left.val : right.val;
if(max.compareTo(node.val) > 0) {
if(left.val.equals(max)) {
left.val = node.val;
node.val = max;
node = left;
} else {
right.val = node.val;
node.val = max;
node = right;
}
} else {
break;
}
} else if(left != null) {
T max = left.val;
if(left.val.compareTo(node.val) > 0) {
left.val = node.val;
node.val = max;
node = left;
} else {
break;
}
} else {
break;
}
} while(true);
}
}
private TreeNode<K> root;
private int position;
PriorityQ(){
this.position = 1;
}
public void insert(K val) {
if(val == null) {
return;
}
if(root == null) {
this.position = 1;
root = new TreeNode<K>(val);
this.position++;
return ;
}
root.insert(val, position);
position++;
}
public K remove() {
if(root == null) {
return null;
}
K val = root.val;
root.remove(this.position-1);
this.position--;
if(position == 1) {
root = null;
}
return val;
}
public static void main(String[] args) {
PriorityQ<Integer> q = new PriorityQ<>();
System.out.println(q.remove());
q.insert(1);
q.insert(11);
q.insert(111);
q.insert(1111);
q.remove();
q.remove();
q.remove();
q.remove();
q.insert(2);
q.insert(4);
}
}