目录

映射map接口

 基于二分搜索树的map

 基于链表的map

 用集合求两个数组的交集

 用映射map求两个数组的交集

• 映射map接口

[code]package com.suanfa.map;
/**
* 映射Map接口
* @author Administrator
*
*/
public interface Map<K,V> {
void add(K key,V value);//向map中添加键值对
V remove(K key);//删除键值对
boolean contains(K key);//判断是否存在某个键和值
V get(K key);//通过键获取值
void set(K key,V value);//修改键值对
int getSize();//获取map大小
boolean isEmpty();//判断map是否为空

}

•  基于二分搜索树的map

[code]package com.suanfa.map;
/**
* 基于二分搜索树的映射map
* @author Administrator
*
*/
public class BSTMap<K extends Comparable<K>,V>implements Map<K,V> {
private class Node{
public K key;
public V value;
public Node left, right;

public Node(K key, V value){
this.key = key;
this.value = value;
left = null;
right = null;
}
}

private Node root;
private int size;

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

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

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

// 向二分搜索树中添加新的元素(key, value)
@Override
public void add(K key, V value){
root = add(root, key, value);
}

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

if(node == null){
size ++;
return new Node(key, value);
}

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

return node;
}

// 返回以node为根节点的二分搜索树中，key所在的节点
private Node getNode(Node node, K key){

if(node == null)
return null;

if(key.equals(node.key))
return node;
else if(key.compareTo(node.key) < 0)
return getNode(node.left, key);
else // if(key.compareTo(node.key) > 0)
return getNode(node.right, key);
}

@Override
public boolean contains(K key){
return getNode(root, key) != null;
}

@Override
public V get(K key){

Node node = getNode(root, key);
return node == null ? null : node.value;
}

@Override
public void set(K key, V newValue){
Node node = getNode(root, key);
if(node == null)
throw new IllegalArgumentException(key + " doesn't exist!");

node.value = newValue;
}

// 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);
}

// 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键为key的节点
@Override
public V remove(K key){

Node node = getNode(root, key);
if(node != null){
root = remove(root, key);
return node.value;
}
return null;
}

private Node remove(Node node, K key){

if( node == null )
return null;

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

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

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

// 待删除节点左右子树均不为空的情况

// 找到比待删除节点大的最小节点, 即待删除节点右子树的最小节点
// 用这个节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;

node.left = node.right = null;

return successor;
}
}

}

•  基于链表的map

[code]package com.suanfa.map;
/**
* 基于链表的映射map
* @author Administrator
*
*/
public class LinkedListMap<K,V>implements Map<K,V> {

private class Node{
public K key;
public V value;
public Node next;

public Node(K key, V value, Node next){
this.key = key;
this.value = value;
this.next = next;
}

public Node(K key, V value){
this(key, value, null);
}

public Node(){
this(null, null, null);
}

@Override
public String toString(){
return key.toString() + " : " + value.toString();
}
}

private Node dummyHead;//虚拟头节点
private int size;//元素个数

public LinkedListMap(){
dummyHead = new Node();
size = 0;
}

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

@Override
public boolean isEmpty(){
return size == 0;
}
/**
* 根据K获取节点
* @param key
* @return
*/
private Node getNode(K key){
Node cur = dummyHead.next;
while(cur != null){
if(cur.key.equals(key))
return cur;
cur = cur.next;
}
return null;
}

@Override
public boolean contains(K key){
return getNode(key) != null;
}

@Override
public V get(K key){
Node node = getNode(key);
return node == null ? null : node.value;
}

@Override
public void add(K key, V value){
Node node = getNode(key);
if(node == null){
dummyHead.next = new Node(key, value, dummyHead.next);
size ++;
}
else
node.value = value;
}

@Override
public void set(K key, V newValue){
Node node = getNode(key);
if(node == null)
throw new IllegalArgumentException(key + " doesn't exist!");

node.value = newValue;
}

@Override
public V remove(K key){

Node prev = dummyHead;
while(prev.next != null){
if(prev.next.key.equals(key))
break;
prev = prev.next;
}

if(prev.next != null){
Node delNode = prev.next;
prev.next = delNode.next;
delNode.next = null;
size --;
return delNode.value;
}

return null;
}

}

•  用集合求两个数组的交集

[code]package com.suanfa.map;

import java.util.ArrayList;
import java.util.TreeSet;

/**
* 用集合求两个数组的交集
* @author Administrator
*
*/
public class Solution349 {
public int[] intersection(int[] nums1, int[] nums2) {

TreeSet<Integer> set = new TreeSet<Integer>();
for(int num: nums1)
set.add(num);

ArrayList<Integer> list = new ArrayList<Integer>();
for(int num: nums2){
if(set.contains(num)){
list.add(num);
set.remove(num);
}
}

int[] res = new int[list.size()];
for(int i = 0 ; i < list.size() ; i ++)
res[i] = list.get(i);
return res;
}

}

•  用映射map求两个数组的交集

[code]package com.suanfa.map;

import java.util.ArrayList;
import java.util.TreeMap;

/**
* 用映射map求两个数组的交集
* @author Administrator
*
*/
public class Solution350 {
public int[] intersect(int[] nums1, int[] nums2) {

TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
for(int num: nums1){
if(!map.containsKey(num))
map.put(num, 1);
else
map.put(num, map.get(num) + 1);
}

ArrayList<Integer> res = new ArrayList<Integer>();
for(int num: nums2){
if(map.containsKey(num)){
res.add(num);
map.put(num, map.get(num) - 1);
if(map.get(num) == 0)
map.remove(num);
}
}

int[] ret = new int[res.size()];
for(int i = 0 ; i < res.size() ; i ++)
ret[i] = res.get(i);

return ret;
}

}