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add solution 146 [solution.java]
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solution/146.Lru Cache/Solution.java

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//双向链表的节点
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class Node{
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public int key;
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public int val;
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public Node pre;//指向前面的指针
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public Node next;//指向后面的指针
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public Node(int key,int value){
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this.val=value;
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this.key=key;
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}
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}
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class LRUCache {
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int capacity;//容量
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Node head;//双向链表的头,维护这个指针,因为set,get时需要在头部操作
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Node end;//双向链表的尾,set时,要是满了,需要将链表的最后一个节点remove
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HashMap<Integer,Node> map=new HashMap<Integer,Node>();//hash表
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public LRUCache(int capacity) {
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this.capacity=capacity;
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}
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//添加,删除尾部,插入头部的操作
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public void remove(Node node){
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Node cur=node;
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Node pre=node.pre;
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Node post=node.next;
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if(pre==null){//说明cur是头部节点
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head=post;
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}
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else pre.next=post;//更新指针,删除
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if(post==null){//说明cur是最后的节点
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end=pre;
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}
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else post.pre=pre;
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}
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public void setHead(Node node){
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//直接插入
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node.next=head;
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node.pre=null;
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if(head!=null) head.pre=node;//防止第一次插入时为空
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head=node;
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if(end==null) end=node;
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}
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public int get(int key) {
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if(map.containsKey(key)){
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//需要把对应的节点调整到头部
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Node latest=map.get(key);
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remove(latest);
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setHead(latest);
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//返回value
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return latest.val;
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}
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else return -1;
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}
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public void put(int key, int value) {
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if(map.containsKey(key)){//这个key原来存在
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//只需要把key对应的node提到最前面,更新value
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Node oldNode=map.get(key);
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oldNode.val=value;
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remove(oldNode);
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setHead(oldNode);
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}
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else{
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//这个key原来不存在,需要重新new出来
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Node newNode=new Node(key,value);
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//接下来要考虑容量
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if(map.size()<capacity){
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setHead(newNode);
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map.put(key,newNode);
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}
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else{
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//容量不够,需要先将map中,最不常使用的那个删除了删除
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map.remove(end.key);
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//接下来更新双向链表
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remove(end);
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setHead(newNode);
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//放入新的
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map.put(key,newNode);
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}
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}
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}
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}
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/**
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* Your LRUCache object will be instantiated and called as such:
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* LRUCache obj = new LRUCache(capacity);
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* int param_1 = obj.get(key);
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* obj.put(key,value);
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*/

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