算法4 Java解答 2.4.24
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2.4.24
问题:
Priority queue with explicit links. Implement a priority queue using a heap- ordered binary tree, but use a triply linked structure instead of an array. You will need three links per node: two to traverse down the tree and one to traverse up the tree. Your implementation should guarantee logarithmic running time per operation, even if no maximum priority-queue size is known ahead of time.
使用链表的优先队列。使用二叉树实现一个优先队列。每个结点都需要三个链接:两个向下,一个向上。你的实现需要保证在无法预知队列大小的情况下也能保证优先队列的基本操作所需要的时间为对数级别。
分析:
插入\删除 :
1、需要寻找对应的位置,时间复杂度为2lgN 2、sink swim 时间复杂度lgN
public class Ex_2_4_24 {
static final class LinkedMaxPQ<Key extends Comparable<Key>>{
Item root;
Item last;
int N;
public void insert(Key v){
Item item = new Item();
item.v = v;
Item prev;
if(last==null){ // 开始堆为空 0 个结点
root = item;
}else if(root == last){ // 已经有1个结点
root.left = item;
item.parent = root;
}else{
prev = last.parent; // 最后一个结点的父节点
if(prev.right==null){ // 非满二叉树,最后一个结点所在的右子树为空
prev.right = item;
item.parent = prev;
}else{ // 最后一个结点所在的右子树已满
while (prev!=root) { // 自底向上,沿着右子树回溯
if (prev != prev.parent.right) {
break;
}
prev = prev.parent;
}
if(prev!=root){ // 非满二叉树:prev为根节点的子树已满
prev = prev.parent.right; // 沿着prev父结点右子树向左下走
}else{ // 满二叉树
prev = root; // 沿着结点左子树一路向下
}
while(prev.left!=null){
prev = prev.left;
}
prev.left = item;
item.parent = prev;
}
}
N++;
last = item;
swim(item);
}
public Key delMax(){
Key max = root.v;
exch(root,last); // 根结点和最后一个结点交换
if(N==1){ // 结点个数为1
root = null;
last = null;
}else if(N==2){ // 结点个数为2
root.left = null;
last = root;
}else{
Item newLast = last;
if(newLast == newLast.parent.right){ // 删除一个结点后,右子树为空,左子树为last结点
newLast = last.parent.left;
}else{
while (newLast!=root){
if(newLast == newLast.parent.right){
break;
}
newLast = newLast.parent;
}
if(newLast==root){ // 删除一个结点后,为满二叉树
}else{
newLast = newLast.parent.left; // 删除一个结点后,为非满二叉树
}
while(newLast.right!=null){
newLast = newLast.right;
}
}
// 删除最后一个结点
if(last.parent.right!=null){
last.parent.right = null;
}else if(last.parent.left!=null){
last.parent.left = null;
}
// 更新最后一个结点
last = newLast;
// 下沉到正确的位置
sink(root);
}
N--;
return max;
}
private void swim(Item v){
while (v.parent!=null && less(v.parent, v)){
exch(v, v.parent);
v = v.parent;
}
}
private void sink(Item v){
Item toExch = null;
while (v.left!=null || v.right!=null){
if(v.left!=null && v.right!=null){ // v存在左右结点
if(less(v.left, v.right)){
toExch = v.right;
}else{
toExch = v.left;
}
}else if(v.left!=null){ // 只有一个结点
toExch = v.left;
}else{
toExch = v.right;
}
if(!less(v,toExch)) break;
exch(v, toExch);
// sink(toExch) // 使用递归也可以
v = toExch;
}
}
private void exch(Item a, Item b){
Key swap = a.v;
a.v = b.v;
b.v = swap;
}
private boolean less(Item a, Item b){
return a.v.compareTo(b.v) < 0;
}
public int size(){
return N;
}
public boolean isEmpty(){
return N==0;
}
public void show(){
show(root);
}
public void show(Item item){
if(item==null) return;
if(item.left==null && item.parent==null) {
return;
}
StdOut.println(item.v.toString());
show(item.left);
show(item.right);
}
public class Item{
Key v;
Item parent;
Item left;
Item right;
@Override
public String toString() {
return "{"+v.toString()+"}";
}
}
}
public static void main(String[] args) {
Integer[] a = randomIntArray(1,10);
LinkedMaxPQ pq = new LinkedMaxPQ();
for(int i=0; i<a.length; i++){
pq.insert(a[i]);
}
pq.show();
StdOut.println("====================================");
for(int i=0; i<a.length; i++){
StdOut.println(pq.delMax());
}
}
//10
//9
//8
//7
//6
//5
//4
//3
//2
//1
}