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LinkedListSet 实现+测试
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liusishan committed Feb 20, 2019
1 parent 6e8393b commit 48fa850
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16,272 changes: 16,272 additions & 0 deletions 07-Set-and-Map/02-LinkedListSet/a-tale-of-two-cities.txt

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13,427 changes: 13,427 additions & 0 deletions 07-Set-and-Map/02-LinkedListSet/pride-and-prejudice.txt

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316 changes: 316 additions & 0 deletions 07-Set-and-Map/02-LinkedListSet/src/BST.java
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import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

/**
* @Auther: lss
* @Date: 2019/1/30 21:40
* @Description:
*/
public class BST<E extends Comparable<E>> {

private class Node {
public E e;
public Node left, right;

public Node(E e) {
this.e = e;
left = null;
right = null;
}

}

private Node root;
private int size;

public BST() {
root = null;
size = 0;
}

public int size() {
return size;
}

public boolean isEmpty() {
return size == 0;
}

// 向二分搜索树中添加新的元素
public void add(E e) {
root = add(root, e);
}

// 向以Node为根的二分搜索树中插入元素E,递归算法
// 返回插入新节点后二分搜索树的根
private Node add(Node node, E e) {

if (node == null) {
size++;
return new Node(e);
}

if (e.compareTo(node.e) < 0)
node.left = add(node.left, e);
else if (e.compareTo(node.e) > 0)
node.right = add(node.right, e);

return node;
}

// 看二分搜索树种是否包含元素e
public boolean contains(E e) {
return contains(root, e);
}

// 看以node为根的二分搜索树种是否包含元素e,递归算法
private boolean contains(Node node, E e) {

if (node == null)
return false;

if (e.compareTo(node.e) == 0)
return true;
else if (e.compareTo(node.e) < 0)
return contains(node.left, e);
else // e.compareTo(node.e) > 0
return contains(node.right, e);
}

// 二分搜索树的前序遍历
public void preOrder() {
preOrder(root);
}

// 前序遍历以node为根的二分搜索树,递归算法
private void preOrder(Node node) {

if (node == null)
return;

System.out.println(node.e);
preOrder(node.left);
preOrder(node.right);
}

// 二分搜索树的非递归前序遍历
public void preOrderNR() {
Stack<Node> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
Node cur = stack.pop();
System.out.println(cur.e);

if (cur.right != null)
stack.push(cur.right);
if (cur.left != null)
stack.push(cur.left);
}
}

// 二分搜索树的中序遍历
public void inOrder() {
inOrder(root);
}

// 中遍历以node 为根的二分搜索树,递归算法
private void inOrder(Node node) {

if (node == null)
return;

inOrder(node.left);
System.out.println(node.e);
inOrder(node.right);
}

// 二分搜索树的后序遍历
public void postOrder() {
postOrder(root);
}

// 后续遍历以node 为根的二分搜索树,递归算法
private void postOrder(Node node) {

if (node == null)
return;

postOrder(node.left);
postOrder(node.right);
System.out.println(node.e);
}

// 二分搜索树的层序遍历
public void levelOrder() {

Queue<Node> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
Node cur = queue.remove();
System.out.println(cur.e);

if (cur.left != null)
queue.add(cur.left);
if (cur.right != null)
queue.add(cur.right);
}
}

// 寻找二分搜索树的最小元素
public E minimum() {
if (size == 0)
throw new IllegalArgumentException("BST is empty");

Node minNode = minimum(root);
return minNode.e;
}

// 返回以node为根的二分搜索树的最小值所在的节点
private Node minimum(Node node) {
if (node.left == null)
return node;
return minimum(node.left);
}

// 寻找二分搜索树的最大元素
public E maximum() {
if (size == 0)
throw new IllegalArgumentException("BST is empty");

Node maxNode = maximum(root);
return maxNode.e;
}

// 返回以node为根的二分搜索树的最大值所在的节点
private Node maximum(Node node) {
if (node.right == null)
return node;
return maximum(node.right);
}

// 从二分搜索树中删除最小值所在节点, 返回最小值
public E removeMin() {
E ret = minimum();
root = removeMin(root);
return ret;
}

// 删除掉以node为根的二分搜索树中的最小节点
// 返回删除节点后新的二分搜索树的根
private Node removeMin(Node node) {

if (node.left == null) {
Node rightNode = node.right;
node.right = null;
size--;
return rightNode;
}

node.left = removeMin(node.left);
return node;
}

// 从二分搜索树中删除最大值所在节点, 返回最大值
public E removeMax() {
E ret = maximum();
root = removeMax(root);
return ret;
}

// 删除掉以node为根的二分搜索树中的最大节点
// 返回删除节点后新的二分搜索树的根
private Node removeMax(Node node) {

if (node.right == null) {
Node leftNode = node.left;
node.left = null;
size--;
return leftNode;
}

node.right = removeMax(node.right);
return node;
}

// 从二分搜索树中删除元素为e的节点
public void remove(E e) {
root = remove(root, e);
}

// 删除以node为根的二分搜索树种值为e的节点,递归算法
// 返回删除节点后新的二分搜索树的根
private Node remove(Node node, E e) {

if (node == null)
return null;

if (e.compareTo(node.e) < 0) {
node.left = remove(node.left, e);
return node;
} else if (e.compareTo(node.e) > 0) {
node.right = remove(node.right, e);
return node;
} else { // e ==node.e

// 待删除结点左子树为空的情况
if (node.left == null) {
Node rightNode = node.right;
node.right = null;
size--;
return rightNode;
}

// 待删除结点右子树为空的情况
if (node.right == null) {
Node leftNode = node.left;
node.left = null;
size--;
return leftNode;
}

// 待删除结点左右子树均不为空的情况
// 找到比待删除结点大的最小结点,即待删除结点右子树的最小结点
// 用这个结点顶替待删除节点的位置
Node successor = minimum(node.right);
successor.right = removeMin(node.right);
successor.left = node.left;
// size++;

node.left = node.right = null;
// size--;

return successor;
}

}

@Override
public String toString() {
StringBuilder res = new StringBuilder();
generateBSTString(root, 0, res);
return res.toString();
}

// 生成以node为根节点,深度为depth的描述二叉树的字符串
private void generateBSTString(Node node, int depth, StringBuilder res) {

if (node == null) {
res.append(generateDepthString(depth) + "null\n");
return;
}

res.append(generateDepthString(depth) + node.e + "\n");
generateBSTString(node.left, depth + 1, res);
generateBSTString(node.right, depth + 1, res);

}

private String generateDepthString(int depth) {
StringBuilder res = new StringBuilder();
for (int i = 0; i < depth; i++)
res.append("--");
return res.toString();
}

}
38 changes: 38 additions & 0 deletions 07-Set-and-Map/02-LinkedListSet/src/BSTSet.java
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/**
* @Auther: lss
* @Date: 2019/2/20 20:08
* @Description:
*/
public class BSTSet<E extends Comparable<E>> implements Set<E> {

private BST<E> bst;

public BSTSet(){
bst = new BST<>();
}

@Override
public void add(E e) {
bst.add(e);
}

@Override
public void remove(E e) {
bst.remove(e);
}

@Override
public boolean contains(E e) {
return bst.contains(e);
}

@Override
public int getSize() {
return bst.size();
}

@Override
public boolean isEmpty() {
return bst.isEmpty();
}
}
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