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LC.P2673[使二叉树所有路径值相等的最小代价] 方法一：贪心 根节点到每个叶子节点的路径值相等，实际上等价于以任意节点为根节点的子树到该子树的每个叶子节点的路径值相等。 12345678910class Solution { public int minIncrements(int n, int[] cost) { int ans = 0; for (int i = n / 2; i > 0; --i) { ans += Math.abs(cost[2 * i - 1] - cost[2 * i]); // 使两个子节点的值变成一样 cost[i - 1] += Math.max(cost[2 * i - 1], cost[2 * i]); // 累加路径和 } return ans; }} 时间复杂度：$O(n)$ 空间复杂度：$O(1)$

LC.P938[二叉搜索树的范围和] 方法一：DFS123456789101112131415161718192021222324252627282930313233/** * 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 i ...

LC.P2583[二叉树中的第K大层和] 方法一：BFS+排序123456789101112131415161718192021222324252627282930313233343536/** * 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 { ...

LC.P889[根据前序和后序遍历构造二叉树] 方法一：DFS+哈希表123456789101112131415161718192021222324252627282930313233343536373839/** * 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 Soluti ...

LC.P106[从中序与后序遍历序列构造二叉树] 方法一：DFS+哈希表12345678910111213141516171819202122232425262728293031323334353637383940/** * 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 Sol ...

LC.P105[从前序与中序遍历序列构造二叉树] 方法一：DFS+哈希表12345678910111213141516171819202122232425262728293031323334353637383940/** * 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 Sol ...

LC.P2641[二叉树的堂兄弟节点II] 方法一：BFS12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849/** * 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; * } * } ...

LCP.P30[魔塔游戏] 方法一：贪心+优先队列123456789101112131415161718192021222324252627class Solution { public int magicTower(int[] nums) { long sum = 0; for (int num : nums) { sum += num; } if (sum < 0) { return -1; } int ans = 0; long hp = 1; PriorityQueue<Integer> q = new PriorityQueue<>(); for (int num : nums) { if (num < 0) { q.offer(num ...

LC.P1686[石子游戏VI] 方法一：贪心+排序123456789101112131415class Solution { public int stoneGameVI(int[] a, int[] b) { int n = a.length; Integer[] index = new Integer[n]; for (int i = 0; i < n; i++) { index[i] = i; } Arrays.sort(index, (i, j) -> a[j] + b[j] - a[i] - b[i]); int diff = 0; for (int i = 0; i < n; i++) { diff += i % 2 == 0 ? a[index[i]] : -b[index[i]]; } return Integer.c ...

CompletableFuture异步多任务最佳实践简介在项目中需要执行异步任务时，通常可以考虑到可以用线程池Executor创建。 如果不需要有返回值，任务实现Runnable接口 如果需要有返回值，任务实现Callable接口，调用Executor的submit方法，再使用Future获取即可 如果多个线程存在依赖组合，可使用同步组件CountDownLatch、CyclicBarrier以及使用起来较为简单的CompeletableFuture等 在Java 8中引入了CompletableFuture，其对Java 5中的Future类提供了非常强大的拓展，可以帮助我们简化异步编程的步骤，并且提供了函数式编程的能力，可以通过回调的方式处理计算结果，也提供了转换和组合CompletableFuture的方法，适用于处理并发的IO密集型任务。 CompletableFuture类图 回顾Future因为CompletableFuture实现了Future接口，因此先回顾Future的用法。 Future是Java 5新加的一个接口，它提供了一种异步并行计算的功能。如果主线程 ...