337. 打家劫舍 III（树形DP）

https://leetcode.cn/problems/house-robber-iii/description/?envType=daily-question&envId=2023-09-18

提示：
树的节点数在 [1, 10^4] 范围内
0 <= Node.val <= 10^4

class Solution {public int rob(TreeNode root) {int[] res = dfs(root);return Math.max(res[0], res[1]);}public int[] dfs(TreeNode root) {// 返回值{a,b} a表示没选当前节点的最大值,b表示选了当前节点的最大值if (root == null) return new int[]{0, 0};int[] l = dfs(root.left), r = dfs(root.right);int a = Math.max(l[0], l[1]) + Math.max(r[0], r[1]), b = root.val + l[0] + r[0];return new int[]{a, b};}
}

2560. 打家劫舍 IV（二分查找+动态规划）

https://leetcode.cn/problems/house-robber-iv/description/?envType=daily-question&envId=2023-09-19

提示：
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
1 <= k <= (nums.length + 1)/2

二分查找答案。 对于每次查找，判断是否可以至少偷k家。

class Solution {public int minCapability(int[] nums, int k) {if (nums.length == 1) return nums[0];int l = Integer.MAX_VALUE, r = Integer.MIN_VALUE;for (int x: nums) {l = Math.min(l, x);r = Math.max(r, x);}// 二分查找答案while (l < r) {int mid = l + r >> 1;if (op(nums, mid) >= k) r = mid;else l = mid + 1;}return l;}// 动态规划public int op(int[] nums, int k) {int n = nums.length;int[] dp = new int[n];      // dp[i]表示0~i中最多能偷几个dp[0] = nums[0] <= k? 1: 0;dp[1] = Math.max(dp[0], nums[1] <= k? 1: 0);for (int i = 2; i < n; ++i) {dp[i] = Math.max(dp[i - 1], dp[i - 2] + (nums[i] <= k? 1: 0));}return dp[n - 1];}
}

LCP 06. 拿硬币（简单贪心模拟）

https://leetcode.cn/problems/na-ying-bi/

class Solution {public int minCount(int[] coins) {int ans = 0;for (int x: coins) ans += (x + 1) / 2;return ans;}
}

2603. 收集树中金币⭐

https://leetcode.cn/problems/collect-coins-in-a-tree/description/?envType=daily-question&envId=2023-09-21

提示：
n == coins.length
1 <= n <= 3 * 10^4
0 <= coins[i] <= 1
edges.length == n - 1
edges[i].length == 2
0 <= ai, bi < n
ai != bi
edges 表示一棵合法的树。

难度分 2712 是因为当时美国站点崩了，很多人没看到题。

思路——拓扑排序删边

https://leetcode.cn/problems/collect-coins-in-a-tree/solutions/2191371/tuo-bu-pai-xu-ji-lu-ru-dui-shi-jian-pyth-6uli/?envType=daily-question&envId=2023-09-21

先去掉所有没有金币的叶子节点。
再去掉最外两层的节点。
最后的答案就是剩余的边数 * 2。

class Solution {public int collectTheCoins(int[] coins, int[][] edges) {int n = coins.length;List<Integer>[] g = new ArrayList[n];Arrays.setAll(g, e -> new ArrayList<>());int[] deg = new int[n];     // 记录每个节点的入度for (int[] e: edges) {int x = e[0], y = e[1];g[x].add(y);g[y].add(x);deg[x]++;deg[y]++;}int leftEdges = n - 1;      // 记录剩余的边数// 拓扑排序，去掉所有没有金币的子树Queue<Integer> q = new LinkedList<>();for (int i = 0; i < n; ++i) {if (deg[i] == 1 && coins[i] == 0) q.offer(i);}while (!q.isEmpty()) {leftEdges--;            // 删除当前节点和其父节点之间的边for (int y: g[q.poll()]) {if (--deg[y] == 1 && coins[y] == 0) {q.offer(y);}}}// 再次拓扑排序，删除最外两层的节点for (int i = 0; i < n; ++i) {if (deg[i] == 1 && coins[i] == 1) q.offer(i);}leftEdges -= q.size();for (int x: q) {for (int y: g[x]) {if (--deg[y] == 1) leftEdges--;}}return Math.max(leftEdges * 2, 0);}
}

2591. 将钱分给最多的儿童（分类讨论）

https://leetcode.cn/problems/distribute-money-to-maximum-children/description/?envType=daily-question&envId=2023-09-22

class Solution {public int distMoney(int money, int children) {if (money < children) return -1;money -= children;int x = Math.min(money / 7, children);      // 计算最多多少个儿童分到8美元int y = money - x * 7;                      // 计算剩余的美元if ((x == children - 1 && y == 3 ) || (x == children  && y > 0)) return x - 1;return x;}
}

1993. 树上的操作💩（设计数据结构）

https://leetcode.cn/problems/operations-on-tree/description/?envType=daily-question&envId=2023-09-23

提示：
n == parent.length
2 <= n <= 2000
对于 i != 0 ，满足 0 <= parent[i] <= n - 1
parent[0] == -1
0 <= num <= n - 1
1 <= user <= 10^4
parent 表示一棵合法的树。
lock ，unlock 和 upgrade 的调用 总共 不超过 2000 次。

class LockingTree {int[] parent;int[] lockNodeUser;List<Integer>[] g;      // 存储所有儿子public LockingTree(int[] parent) {int n = parent.length;this.parent = parent;lockNodeUser = new int[n];Arrays.fill(lockNodeUser, -1);g = new List[n];Arrays.setAll(g, e -> new ArrayList<>());for (int i = 0; i < n; ++i) {if (parent[i] != -1) g[parent[i]].add(i);}}public boolean lock(int num, int user) {if (lockNodeUser[num] == -1) {lockNodeUser[num] = user;return true;}return false;}public boolean unlock(int num, int user) {if (lockNodeUser[num] == user) {lockNodeUser[num] = -1;return true;}return false;}public boolean upgrade(int num, int user) {// 自己没被上锁，没有祖宗上锁，有子孙节点上锁了boolean res = lockNodeUser[num] == -1 && !hasLockedAncestor(num) && checkAndUnlockDescendant(num);if (res) lockNodeUser[num] = user;return res;}// 是否有祖宗节点被上锁public boolean hasLockedAncestor(int num) {num = parent[num];while (num != -1) {if (lockNodeUser[num] != -1) return true;num = parent[num];}return false;}// 是否有子孙节点被上锁，并解锁public boolean checkAndUnlockDescendant(int num) {boolean res = lockNodeUser[num] != -1;lockNodeUser[num] = -1;         for (int y: g[num]) {res |= checkAndUnlockDescendant(y);}return res;}
}/*** Your LockingTree object will be instantiated and called as such:* LockingTree obj = new LockingTree(parent);* boolean param_1 = obj.lock(num,user);* boolean param_2 = obj.unlock(num,user);* boolean param_3 = obj.upgrade(num,user);*/

这题的重点在于操作三的实现。

146. LRU 缓存（⭐数据结构：哈希表+双向链表）

https://leetcode.cn/problems/lru-cache/description/?envType=daily-question&envId=2023-09-24

提示：

1 <= capacity <= 3000
0 <= key <= 10000
0 <= value <= 10^5
最多调用 2 * 10^5 次 get 和 put

解法1——哈希表+双向链表⭐

双向链表维护各个节点被使用的情况，头节点是最近被使用的，尾节点是最久未被使用的。
哈希表维护key和节点之间的映射，帮助快速找到指定key的节点。

class LRUCache {class DLinkedNode {int key;int value;DLinkedNode prev;DLinkedNode next;public DLinkedNode() {};public DLinkedNode(int _key, int _value) {this.key = _key;this.value = _value;}}Map<Integer, DLinkedNode> cache = new HashMap<>();  // key和节点的映射int size = 0;       // 大小int capacity;       // 容量// 虚拟头尾节点DLinkedNode head = new DLinkedNode(), tail = new DLinkedNode();public LRUCache(int capacity) {this.capacity = capacity;head.next = tail;tail.prev = head;}public int get(int key) {DLinkedNode node = cache.get(key);if (node == null) return -1;moveToHead(node);return node.value;}public void put(int key, int value) {DLinkedNode node = cache.get(key);if (node == null) {DLinkedNode newNode = new DLinkedNode(key, value);cache.put(key, newNode);addToHead(newNode);++size;if (size > capacity) {DLinkedNode last = removeTail();    cache.remove(last.key);--size;} } else {node.value = value;moveToHead(node);}}// 将节点添加到头部public void addToHead(DLinkedNode node) {node.prev = head;node.next = head.next;head.next.prev = node;head.next = node;}// 删除节点public void removeNode(DLinkedNode node) {node.prev.next = node.next;node.next.prev = node.prev;}// 将节点移动到头部public void moveToHead(DLinkedNode node) {removeNode(node);addToHead(node);}// 删除最后一个节点public DLinkedNode removeTail() {DLinkedNode node = tail.prev;removeNode(node);return node;}
}/*** Your LRUCache object will be instantiated and called as such:* LRUCache obj = new LRUCache(capacity);* int param_1 = obj.get(key);* obj.put(key,value);*/

解法2——Java JDK LinkedHashMap

class LRUCache extends LinkedHashMap<Integer, Integer>{private int capacity;public LRUCache(int capacity) {super(capacity, 0.75F, true);this.capacity = capacity;}public int get(int key) {return super.getOrDefault(key, -1);}public void put(int key, int value) {super.put(key, value);}@Overrideprotected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {return size() > capacity;} 
}

补充——LinkedHashMap

https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/util/LinkedHashMap.html

构造器：

---

protected boolean removeEldestEntry​(Map.Entry<K,​V> eldest)
如果此映射应该删除其最年长的条目，则返回true。在向映射中插入新条目后，put和putAll调用该方法。它为实现者提供了每次添加新条目时删除最老条目的机会。如果映射表示缓存，这很有用:它允许映射通过删除过时的条目来减少内存消耗。

补充——Java修饰符