-
Preface
- FAQ
-
Part I - Basics
- Basics Data Structure
- Basics Sorting
- Basics Algorithm
- Basics Misc
-
Part II - Coding
- String
-
Integer Array
-
Remove Element
-
Zero Sum Subarray
-
Subarray Sum K
-
Subarray Sum Closest
-
Recover Rotated Sorted Array
-
Product of Array Exclude Itself
-
Partition Array
-
First Missing Positive
-
2 Sum
-
3 Sum
-
3 Sum Closest
-
Remove Duplicates from Sorted Array
-
Remove Duplicates from Sorted Array II
-
Merge Sorted Array
-
Merge Sorted Array II
-
Median
-
Partition Array by Odd and Even
-
Kth Largest Element
-
Remove Element
-
Binary Search
-
First Position of Target
-
Search Insert Position
-
Search for a Range
-
First Bad Version
-
Search a 2D Matrix
-
Search a 2D Matrix II
-
Find Peak Element
-
Search in Rotated Sorted Array
-
Search in Rotated Sorted Array II
-
Find Minimum in Rotated Sorted Array
-
Find Minimum in Rotated Sorted Array II
-
Median of two Sorted Arrays
-
Sqrt x
-
Wood Cut
-
First Position of Target
-
Math and Bit Manipulation
-
Single Number
-
Single Number II
-
Single Number III
-
O1 Check Power of 2
-
Convert Integer A to Integer B
-
Factorial Trailing Zeroes
-
Unique Binary Search Trees
-
Update Bits
-
Fast Power
-
Hash Function
-
Happy Number
-
Count 1 in Binary
-
Fibonacci
-
A plus B Problem
-
Print Numbers by Recursion
-
Majority Number
-
Majority Number II
-
Majority Number III
-
Digit Counts
-
Ugly Number
-
Plus One
-
Palindrome Number
-
Task Scheduler
-
Single Number
-
Linked List
-
Remove Duplicates from Sorted List
-
Remove Duplicates from Sorted List II
-
Remove Duplicates from Unsorted List
-
Partition List
-
Add Two Numbers
-
Two Lists Sum Advanced
-
Remove Nth Node From End of List
-
Linked List Cycle
-
Linked List Cycle II
-
Reverse Linked List
-
Reverse Linked List II
-
Merge Two Sorted Lists
-
Merge k Sorted Lists
-
Reorder List
-
Copy List with Random Pointer
-
Sort List
-
Insertion Sort List
-
Palindrome Linked List
-
LRU Cache
-
Rotate List
-
Swap Nodes in Pairs
-
Remove Linked List Elements
-
Remove Duplicates from Sorted List
-
Binary Tree
-
Binary Tree Preorder Traversal
-
Binary Tree Inorder Traversal
-
Binary Tree Postorder Traversal
-
Binary Tree Level Order Traversal
-
Binary Tree Level Order Traversal II
-
Maximum Depth of Binary Tree
-
Balanced Binary Tree
-
Binary Tree Maximum Path Sum
-
Lowest Common Ancestor
-
Invert Binary Tree
-
Diameter of a Binary Tree
-
Construct Binary Tree from Preorder and Inorder Traversal
-
Construct Binary Tree from Inorder and Postorder Traversal
-
Subtree
-
Binary Tree Zigzag Level Order Traversal
-
Binary Tree Serialization
-
Binary Tree Preorder Traversal
- Binary Search Tree
- Exhaustive Search
-
Dynamic Programming
-
Triangle
-
Backpack
-
Backpack II
-
Minimum Path Sum
-
Unique Paths
-
Unique Paths II
-
Climbing Stairs
-
Jump Game
-
Word Break
-
Longest Increasing Subsequence
-
Palindrome Partitioning II
-
Longest Common Subsequence
-
Edit Distance
-
Jump Game II
-
Best Time to Buy and Sell Stock
-
Best Time to Buy and Sell Stock II
-
Best Time to Buy and Sell Stock III
-
Best Time to Buy and Sell Stock IV
-
Distinct Subsequences
-
Interleaving String
-
Maximum Subarray
-
Maximum Subarray II
-
Longest Increasing Continuous subsequence
-
Longest Increasing Continuous subsequence II
-
Maximal Square
-
Triangle
- Graph
- Data Structure
- Big Data
- Problem Misc
-
Part III - Contest
- Google APAC
- Microsoft
- Appendix I Interview and Resume
-
Tags
LRU Cache
Question
- leetcode: LRU Cache | LeetCode OJ
- lintcode: (134) LRU Cache
Problem Statement
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get
and set
.
get(key)
- Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(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.
题解
Java
public class Solution {
private int capacity;
private HashMap<Integer, Node> map = new HashMap<>();
private Node head = new Node(-1, -1), tail = new Node(-1, -1);
private class Node {
Node prev, next;
int val, key;
public Node(int key, int val) {
this.val = val;
this.key = key;
prev = null;
next = null;
}
// @Override
// public String toString() {
// return "(" + key + ", " + val + ") " + "last:"
// + (prev == null ? "null" : "node");
// }
}
public Solution(int capacity) {
this.capacity = capacity;
tail.prev = head;
head.next = tail;
}
public int get(int key) {
if (!map.containsKey(key)) {
return -1;
}
// remove current
Node currentNode = map.get(key);
currentNode.prev.next = currentNode.next;
currentNode.next.prev = currentNode.prev;
// move current to tail;
moveToTail(currentNode);
return map.get(key).val;
}
public void set(int key, int value) {
if (get(key) != -1) {
map.get(key).val = value;
return;
}
if (map.size() == capacity) {
map.remove(head.next.key);
head.next = head.next.next;
head.next.prev = head;
}
Node insert = new Node(key, value);
map.put(key, insert);
moveToTail(insert);
}
private void moveToTail(Node current) {
current.prev = tail.prev;
tail.prev = current;
current.prev.next = current;
current.next = tail;
}
}
copy