一.问题描述 昨天晚上Myeclipse还用着好好的,今天早上打开工程,只要运行就卡住,大半天弹出个消息窗口:Java(TM) Platform SE binary 已停止工作. 如图 关闭Myeclipse之后出现java was started but returned exit code=805306369 如图 二.解决过程 1.改了Myeclipse的ini配置文件 把MyEclipse启动配置文件ini中找到Vm这一行.自己安装的java虚拟机 -vmC:/Program Files
Language Support for Java(TM) by Red Hat(1.3.0) 注意:版本问题,可能会有部分出入 功能目录 设置 java.home 作用: 指定用于启动 Java 语言服务器的 JDK (11或更近版本)的文件夹路径.在 Windows 上,反斜杠必须转义 "java.home":"C:\\Program Files\\Java\\jdk11.0_8" 默认值:null java.jdt.ls.java.home 作用: 指定用于启
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order t
错误描述:Java 安装时断电,再次安装java时,提示“您的电脑上已经安装了此软件.是否要重新安装”,点“是”后出现“内部错误2753:RegUtils”,点“确定”又出现上述提示. 解决办法 :使用Windows Install Clean Up 将 Java SE Development Kit 删除即可.
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example: Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,
Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. 后序遍历,左子树→右子树→根节点 前序遍历的非递归实现需要一个计数器,方法是需要重写一个类继承TreeNode,翁慧玉教材<数据结构:题解与拓展>P113有详细介绍,这里略.递归JAVA实现如下: public L
Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 二叉树的前序遍历,根节点→左子树→右子树 解题思路一: 递归实现,JAVA实现如下: public List<Integer> preorderTraversal(TreeNode root) { List<I
Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 Return 6. 解题思路: DFS暴力枚举,注意,如果采用static 全局变量的话,在IDE里面是可以通过,但在OJ上无法测试通过,因此需要建立一个类来储存结果,JAVA实现如下: publ
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7] ] 解