算法-刷题

桃扇骨 2022-09-09 02:26 222阅读 0赞

## 剑指Offer快速过了一遍 ##

## 字节-飞书高频题-前15 ##

各公司的高频面试算法题，可在 [CodeTop][]查询

#### (1) 146. LRU 缓存机制 ####

**我的实现：**

class LRUCache {
        
    
        ArrayDeque<Integer> queue;
        Map<Integer, Integer> cacheMap;
        private int capacity;
        private int size;
        public LRUCache(int capacity) {
        
            if (capacity < 1) {
        
                throw new RuntimeException("capacity is error");
            }
            this.capacity = capacity;
            queue = new ArrayDeque<>();
            cacheMap = new HashMap<>();
            size = 0;
    
        }
    
        public int get(int key) {
        
            if (cacheMap.isEmpty()) {
        
                return -1;
            }
            if (cacheMap.containsKey(key)) {
        
                queue.remove(key);
                queue.add(key);
                return cacheMap.get(key);
            } else {
        
                return -1;
            }
        }
    
       public void put(int key, int value) {
        
            if (cacheMap.containsKey(key)) {
        
                cacheMap.put(key, value);
                queue.remove(key);
                queue.add(key);
            } else {
        
                size ++;
                if (size > capacity) {
        
                    int needDelete = queue.pop();
                    cacheMap.remove(needDelete);
                }
                cacheMap.put(key, value);
                queue.add(key);
            }
        }
    }

**官方实现：**

public class LRUCache {
        
        class DLinkedNode {
        
            int key;
            int value;
            DLinkedNode prev;
            DLinkedNode next;
            public DLinkedNode() {
        }
            public DLinkedNode(int _key, int _value) {
        key = _key; value = _value;}
        }
    
        private Map<Integer, DLinkedNode> cache = new HashMap<Integer, DLinkedNode>();
        private int size;
        private int capacity;
        private DLinkedNode head, tail;
    
        public LRUCache(int capacity) {
        
            this.size = 0;
            this.capacity = capacity;
            // 使用伪头部和伪尾部节点
            head = new DLinkedNode();
            tail = new DLinkedNode();
            head.next = tail;
            tail.prev = head;
        }
    
        public int get(int key) {
        
            DLinkedNode node = cache.get(key);
            if (node == null) {
        
                return -1;
            }
            // 如果 key 存在，先通过哈希表定位，再移到头部
            moveToHead(node);
            return node.value;
        }
    
        public void put(int key, int value) {
        
            DLinkedNode node = cache.get(key);
            if (node == null) {
        
                // 如果 key 不存在，创建一个新的节点
                DLinkedNode newNode = new DLinkedNode(key, value);
                // 添加进哈希表
                cache.put(key, newNode);
                // 添加至双向链表的头部
                addToHead(newNode);
                ++size;
                if (size > capacity) {
        
                    // 如果超出容量，删除双向链表的尾部节点
                    DLinkedNode tail = removeTail();
                    // 删除哈希表中对应的项
                    cache.remove(tail.key);
                    --size;
                }
            }
            else {
        
                // 如果 key 存在，先通过哈希表定位，再修改 value，并移到头部
                node.value = value;
                moveToHead(node);
            }
        }
    
        private void addToHead(DLinkedNode node) {
        
            node.prev = head;
            node.next = head.next;
            head.next.prev = node;
            head.next = node;
        }
    
        private void removeNode(DLinkedNode node) {
        
            node.prev.next = node.next;
            node.next.prev = node.prev;
        }
    
        private void moveToHead(DLinkedNode node) {
        
            removeNode(node);
            addToHead(node);
        }
    
        private DLinkedNode removeTail() {
        
            DLinkedNode res = tail.prev;
            removeNode(res);
            return res;
        }
    }
    
    作者：LeetCode-Solution
    链接：https://leetcode-cn.com/problems/lru-cache/solution/lruhuan-cun-ji-zhi-by-leetcode-solution/
    来源：力扣（LeetCode）

#### (2) 3. 无重复字符的最长子串 ####

class Solution {
        
        public int lengthOfLongestSubstring(String s) {
        
            if (s == null || s.length() == 0) {
        
                return 0;
            }
            int max = 0;
            int left = 0;
            HashMap<Character, Integer> map = new HashMap<>();
            for (int i = 0; i< s.length(); i++ ) {
        
                Character charTemp = s.charAt(i);
                if (map.containsKey(charTemp)) {
        
                    left = Math.max(left, map.get(charTemp) + 1);
                }
                map.put(charTemp, i);
                max = Math.max(max, i - left + 1);
            }
            return max;
        }
    }

#### (3) 206. 反转链表 ####

/**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode() {}
     *     ListNode(int val) { this.val = val; }
     *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
     * }
     */
    class Solution {
        
        public ListNode reverseList(ListNode head) {
        
            if (head == null || head.next == null) {
         
                return head;
            }
            ListNode preNode = null;
            ListNode currentNode = head;
            while (currentNode != null) {
        
                ListNode next = currentNode.next;
                currentNode.next = preNode;
                preNode = currentNode;
                currentNode = next;
            }
            return preNode;
        }
    }

#### (4) 103. 二叉树的锯齿形层序遍历 ####

/**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode() {}
     *     TreeNode(int val) { this.val = val; }
     *     TreeNode(int val, TreeNode left, TreeNode right) {
     *         this.val = val;
     *         this.left = left;
     *         this.right = right;
     *     }
     * }
     */
    class Solution {
        
        public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        
            List<List<Integer>> result = new LinkedList<List<Integer>>();
            if (root == null) {
        
                return result;
            }
            Deque<TreeNode> deque = new LinkedList<>();
            deque.offer(root);
            int currentLevel = 0;
            while (!deque.isEmpty()) {
        
                int levelSize = deque.size();
                Deque<Integer> currentList = new LinkedList<>();
                for (int i=0; i< levelSize; i++) {
        
                    TreeNode currentNode = deque.peek();
                    if (currentNode.left != null) {
        
                        deque.offer(currentNode.left);
                    }
                    if (currentNode.right != null) {
        
                        deque.offer(currentNode.right);
                    }
                    if (currentLevel % 2 == 0) {
        
                        currentList.offerLast(currentNode.val);
                    } else {
        
                        currentList.offerFirst(currentNode.val);
                    }
                    deque.poll();
                }
                currentLevel ++ ;
                result.add(new LinkedList<>(currentList));
            }
            return result;
          
        }
    }

该题是 102.二叉树的层序遍历 的变换之后的题目，下面是 二叉树层序遍历的实现：

class Solution {
        
        public List<List<Integer>> levelOrder(TreeNode root) {
        
           List<List<Integer>> result = new LinkedList<List<Integer>>();
            if (root == null) {
        
                return result;
            }
            Deque<TreeNode> deque = new LinkedList<TreeNode>();
            deque.offer(root);
            while (!deque.isEmpty()) {
        
                int size = deque.size();
                List<Integer> currentList = new LinkedList<>();
                for (int i=0; i< size; i++) {
        
                    TreeNode currentNode = deque.peek();
                    currentList.add(currentNode.val);
                    if (currentNode.left != null) {
        
                        deque.offer(currentNode.left);
                    }
                    if (currentNode.right != null) {
        
                        deque.offer(currentNode.right);
                    }
                    deque.poll();
                }
                result.add(currentList);
            }
            return result;
        }
    }

#### (5) 121.买卖股票的最佳时机 ####

public class Solution {
        
        /**
         *暴力解法
         * @param prices
         * @return
         */
        public int maxProfit1(int[] prices) {
        
            int maxProfit = 0;
            for (int i = 0; i<prices.length - 1; i++) {
        
                for (int j = i+1; j < prices.length; j++) {
        
                    int currentProfit = prices[j] - prices[i];
                    if (currentProfit > maxProfit) {
        
                        maxProfit = currentProfit;
                    }
                }
            }
            return maxProfit;
        }
    
        /**
         * 非暴力
         * @param prices
         * @return
     
        */
        public int maxProfit(int[] prices) {
        
            int minPrice = Integer.MAX_VALUE;
            int maxProfit = 0;
            for (int i = 0; i<prices.length; i++) {
        
               if (prices[i] < minPrice) {
        
                   minPrice = prices[i];
               } else if (prices[i] - minPrice > maxProfit) {
        
                   maxProfit = prices[i] - minPrice;
               }
            }
            return maxProfit;
        }
    
    }

#### (6) 41.缺失的第一个正数 ####

class Solution {
        
        public int firstMissingPositive(int[] nums) {
        
            for (int i = 0; i<nums.length;i++) {
        
                while (nums[i] >=1 && nums[i]<=nums.length && nums[i] != nums[nums[i] -1]) {
        
                    swap(nums, i, nums[i] -1);
                }
            }
            
            for (int i=0;i<nums.length;i++) {
        
                if (i != nums[i]-1) {
        
                    return i+1;
                }
            }
            return nums.length +1;
        }
    
    
        private void swap(int[] nums, int i,int j) {
        
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
        }
    }

#### (7) 46.全排列 ####

public class Permute {
        
    
        public List<List<Integer>> permute(int[] nums) {
        
            List<List<Integer>> result = new ArrayList<>();
            if (nums == null || nums.length == 0) {
        
                return result;
            }
            List<Integer> outPut = new ArrayList<>();
            for (int i=0; i< nums.length; i++) {
        
                outPut.add(nums[i]);
            }
            backTrack(0, nums.length, outPut, result);
            return result;
        }
    
        public void backTrack(int first, int n, List<Integer> outPut, List<List<Integer>> result) {
        
            if (first == n) {
        
                result.add(new ArrayList<>(outPut));
            }
            for (int i = first; i<n; i++) {
        
                Collections.swap(outPut, first, i);
                backTrack(first+1, n, outPut, result);
                Collections.swap(outPut, first, i);
            }
        }
    }

#### (8) 1. 两数之和 ####

class Solution {
        
        public int[] twoSum(int[] nums, int target) {
        
            HashMap<Integer, Integer> numAndIndex = new HashMap<>();
            for (int i=0;i<nums.length;i++) {
        
                int b = target - nums[i];
                if (numAndIndex.containsKey(b)) {
        
                    return new int[]{
        i, numAndIndex.get(b)};
                }
                numAndIndex.put(nums[i], i);
            }
            return new int[0];
        }  
    }

#### (9) 128. 最长连续序列 ####

public int longestConsecutive(int[] nums) {
        
            HashSet<Integer> numSet = new HashSet<>();
            for (int i=0; i<nums.length;i++) {
        
                numSet.add(nums[i]);
            }
            int maxLength = 0;
            for (int i =0;i<nums.length;i++) {
        
                if (!numSet.contains(nums[i] - 1)) {
        
                    int currentNum = nums[i];
                    int currentLength = 1;
                    while (numSet.contains(currentNum + 1)) {
        
                        currentNum +=1;
                        currentLength ++;
                    }
                    maxLength = Math.max(maxLength, currentLength);
                }
            }
            return maxLength;
        }

[CodeTop]: https://codetop.cc/home