打印二叉堆:利用层级关系
我这里是先将堆排序,然后在sort里执行了打印堆的方法printAsTree()
public class MaxHeap<T extends Comparable<? super T>> { private T[] data; private int size; private int capacity; public MaxHeap(int capacity) { this.capacity = capacity; this.size = 0; this.data = (T[]) new Comparable[capacity + 1]; } public MaxHeap(T[] arr) {//heapify,数组建堆 capacity = arr.length; data = (T[]) new Comparable[capacity + 1]; System.arraycopy(arr, 0, data, 1, arr.length); size = arr.length; for (int i = size / 2; i >= 1; i--) { shiftDown(i); } } public int size() { return this.size; } public int getCapacity() { return this.capacity; } public boolean isEmpty() { return size == 0; } public T seekMax() { return data[1]; } public void swap(int i, int j) { if (i != j) { T temp = data[i]; data[i] = data[j]; data[j] = temp; } } public void insert(T item) { size++; data[size] = item; shiftUp(size); } public T popMax() { swap(1, size--); shiftDown(1); return data[size + 1]; } public void shiftUp(int child) { while (child > 1 && data[child].compareTo(data[child / 2]) > 0) { swap(child, child / 2); child /= 2; } } /** * @param a data数组中某个元素的下角标 * @param b data数组中某个元素的下角标 * @return 哪个元素大就返回哪个的下角标 */ private int max(int a, int b) { if (data[a].compareTo(data[b]) < 0) {//如果data[b]大 return b;//返回b } else {//如果data[a]大 return a;//返回a } } /** * @param a data数组中某个元素的下角标 * @param b data数组中某个元素的下角标 * @param c data数组中某个元素的下角标 * @return 哪个元素大就返回哪个的下角标 */ private int max(int a, int b, int c) { int biggest = max(a, b); biggest = max(biggest, c); return biggest; } public void shiftDown(int father) { while (true) { int lchild = father * 2; int rchild = father * 2 + 1; int newFather = father;//这里赋不赋值无所谓,如果把下面这个return改成break,那就必须赋值了 if (lchild > size) {//如果没有左、右孩子 return; } else if (rchild > size) {//如果没有右孩子 newFather = max(father, lchild); } else {//如果有左、右孩子 newFather = max(father, lchild, rchild); } if (newFather == father) {//如果原父结点就是三者最大,则不用继续整理堆了 return; } else {//父节点不是最大,则把大的孩子交换上来,然后继续往下堆调整,直到满足大根堆为止 swap(newFather, father); father = newFather;//相当于继续shiftDown(newFather)。假如newFather原来是father的左孩子,那就相当于shiftDown(2*father) } } } public static <T extends Comparable<? super T>> void sort(T[] arr) { int len = arr.length; MaxHeap<T> maxHeap = new MaxHeap<>(arr); maxHeap.printAsTree(); for (int i = len - 1; i >= 0; i--) { arr[i] = maxHeap.popMax(); } } public static void printArr(Object[] arr) { for (Object o : arr) { System.out.print(o); System.out.print("\t"); } System.out.println(); } public void printSpace(int n) {//打印n个空格(在这里用‘\t’来代替) for (int i = 0; i < n; i++) { System.out.printf("%3s", ""); } } public void printAsTree() { int lineNum = 1;//首先遍历第一行 int lines = (int) (Math.log(size) / Math.log(2)) + 1;//lines是堆的层数 int spaceNum = (int) (Math.pow(2, lines) - 1); for (int i = 1; i <= size; ) { //因为在[1...size]左闭右闭区间存数据,data[0]不存数据 //每层都是打印这个区间[2^(层数-1) ... (2^层数)-1]。如果堆里的数不够(2^层数)-1个,那就打印到size。所以取min((2^层数)-1,size). for (int j = (int) Math.pow(2, lineNum - 1); j <= Math.min(size, (int) Math.pow(2, lineNum) - 1); j++) { printSpace(spaceNum); //打印spaceNum个空格 System.out.printf("%3s", data[j]);//打印数据 System.out.printf("%3s", "");//图片中绿色方框 printSpace(spaceNum);//打印spaceNum个空格 i++;//每打印一个元素就 + 1 } lineNum++; spaceNum = spaceNum / 2; System.out.println(); } } public static void main(String args[]) { Integer[] arr = {3, 5, 1, 7, 2, 9, 8, 0, 4, 6, 1, 3, 6, 1, 1}; sort(arr); } }
执行结果:
时间: 2024-10-07 07:36:01