[LeetCode] 146. LRU Cache_Hard tag: Hash, Linked List

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

这个题目实际上就是用linked list去记录cache中存在的点，因为delete和move，add都是O(1) 的复杂度，然后用hash 去存cache中现在有的数。

Note：因为存的是key，所以linked list里面还多一个key的值。另外由于如果用单向的linked list，我们要知道它的pre node，所以在hash中我们存的实际上是该node的pre node，因此也需要有个dummy node。另外因为有去掉head，add to tail的操作，所以可以另写两个function，有助于debug，也使程序看起来更加有条理。

Code:

class ListNode:
    def __init__(self, key, value):
        self.key = key
        self.val = value
        self.next = None

class LRUCache:
    def __init__(self, capacity):
        self.capacity = capacity
        self.d = dict()  # hash
        self.dummy = ListNode(0, 0)
        self.tail = self.dummy
    def popHead(self):
        head = self.dummy.next
        self.dummy.next = head.next
        self.d.pop(head.key)
        if head.next:
            self.d[head.next.key] = self.dummy
    def addToTail(self, node):
        self.tail.next = node
        self.d[node.key] = self.tail
        self.tail = node
    def moveToTail(self, pre):
        if pre.next == self.tail:
            return 
        node = pre.next
        pre.next = node.next
        if node.next:
            self.d[node.next.key] = pre
            node.next = None
        self.addToTail(node)
    def get(self, key):
        if key in self.d:
            self.moveToTail(self.d[key])
            return self.d[key].next.val
        return -1
    def put(self, key, value):
        if key in self.d:
            self.moveToTail(self.d[key])
            self.d[key].next.val = value
        else:
            self.addToTail(ListNode(key, value))
            if len(self.d) > self.capacity:
                self.popHead()