ug4.0启动服务不能安装怎么办

1、先安装java程序，

2、修改许可许可文件。在\MAGNiTUDE\win32下面有个ugs4lic文件。修改其中的this_host为你电脑计算机的名字
3首先安装UG的许可服务器。打开安装软件，找到ugslicensing040这个文件夹，打开。里面有个setupexe。双击进行安装。
4复制\MAGNiTUDE\win32下的ugslmdexe到你刚刚安装的许可服务器路径下面。覆盖掉原文件。
5安装UG的主程序，在nx075这个文件夹下面有个setupexe，双击进行安装。
6破解。把\MAGNiTUDE\win32\ProgramFiles\UGS\NX 75 这个文件夹下面的5个文件夹。复制到你的UG安装目录下。复制过程中点“全部”。
7之后打开UG软件，会出现许可服务器错误。
8运行LMTOOLS配置下如图

安装完成，重启电脑。

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楼主的情况是不是jdk安装不上，但jre可以？如果是的话，就是以前安装的jdk卸载不干净，楼主可以进入注册表，搜索java，把关于java的注册表都删了，然后在安装就行了，不一定能帮上忙，楼主可以试试

前端刷题用js还是java
前端刷题用js还是java_用JavaScript刷LeetCode的正确姿势

韦桂超
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虽然很多人都觉得前端算法弱，但其实 JavaScript 也可以刷题啊！最近两个月断断续续刷完了 leetcode 前 200 的 middle + hard ，总结了一些刷题常用的模板代码。走过路过发现 bug 请指出，拯救一个辣鸡(但很帅)的少年就靠您啦！
常用函数
包括打印函数和一些数学函数。
const _max =Mathmaxbind(Math);
const _min=Mathminbind(Math);
const _pow=Mathpowbind(Math);
const _floor=Mathfloorbind(Math);
const _round=Mathroundbind(Math);
const _ceil=Mathceilbind(Math);
const log=consolelogbind(console);//const log = _ => {}
log 在提交的代码中当然是用不到的，不过在调试时十分有用。但是当代码里面加了很多 log 的时候，提交时还需要一个个注释掉就相当麻烦了，只要将 log 赋值为空函数就可以了。
举一个简单的例子，下面的代码是可以直接提交的。
//计算 1+2++n//const log = consolelogbind(console);
const log = _ =>{}functionsumOneToN(n) {
let sum= 0;for (let i = 1; i <= n; i++) {
sum+=i;
log(`i=${i}: sum=${sum}`);
}returnsum;
}
sumOneToN(10);
位运算的一些小技巧
判断一个整数 x 的奇偶性： x & 1 = 1 (奇数) ， x & 1 = 0 (偶数)
求一个浮点数 x 的整数部分： ~~x ，对于正数相当于 floor(x) 对于负数相当于 ceil(-x)
计算 2 ^ n ： 1 << n 相当于 pow(2, n)
计算一个数 x 除以 2 的 n 倍： x >> n 相当于 ~~(x / pow(2, n))
判断一个数 x 是 2 的整数幂(即 x = 2 ^ n ): x & (x - 1) = 0
※注意※：上面的位运算只对32位带符号的整数有效，如果使用的话，一定要注意数！据！范！围！
记住这些技巧的作用：
提升运行速度 ❌
提升逼格 ✅
举一个实用的例子，快速幂(原理自行google)
//计算x^n n为整数
functionqPow(x, n) {
let result= 1;while(n) {if (n & 1) result = x; //同 if(n%2)
x = x x;
n>>= 1; //同 n=floor(n/2)
}returnresult;
}
链表
刚开始做 LeetCode 的题就遇到了很多链表的题。恶心心。最麻烦的不是写题，是调试啊！！于是总结了一些链表的辅助函数。
/
链表节点
@param {} val
@param {ListNode} next/
function ListNode(val, next = null) {thisval =val;thisnext =next;
}/
将一个数组转为链表
@param {array} a
@return {ListNode}/const getListFromArray= (a) =>{
let dummy= newListNode()
let pre=dummy;
aforEach(x=> pre = prenext = newListNode(x));returndummynext;
}/
将一个链表转为数组
@param {ListNode} node
@return {array}/const getArrayFromList= (node) =>{
let a=[];while(node) {
apush(nodeval);
node=nodenext;
}returna;
}/
打印一个链表
@param {ListNode} node/const logList= (node) =>{
let str= 'list: ';while(node) {
str+= nodeval + '->';
node=nodenext;
}
str+= 'end';
log(str);
}
还有一个常用小技巧，每次写链表的 *** 作，都要注意判断表头，如果创建一个空表头来进行 *** 作会方便很多。
let dummy = newListNode();//返回
return dummynext;
使用起来超爽哒~举个例子。@leetcode 82。题意就是删除链表中连续相同值的节点。
/ @lc app=leetcode id=82 lang=javascript

[82] Remove Duplicates from Sorted List II/
/
@param {ListNode} head
@return {ListNode}/
var deleteDuplicates = function(head) {//空指针或者只有一个节点不需要处理
if (head === null || headnext === null) returnhead;
let dummy= newListNode();
let oldLinkCurrent=head;
let newLinkCurrent=dummy;while(oldLinkCurrent) {
let next=oldLinkCurrentnext;//如果当前节点和下一个节点的值相同 就要一直向前直到出现不同的值
if (next && oldLinkCurrentval ===nextval) {while (next && oldLinkCurrentval ===nextval) {
next=nextnext;
}
oldLinkCurrent=next;
}else{
newLinkCurrent= newLinkCurrentnext =oldLinkCurrent;
oldLinkCurrent=oldLinkCurrentnext;
}
}
newLinkCurrentnext= null; //记得结尾置空~
logList(dummynext);returndummynext;
};
deleteDuplicates(getListFromArray([1,2,3,3,4,4,5]));
deleteDuplicates(getListFromArray([1,1,2,2,3,3,4,4,5]));
deleteDuplicates(getListFromArray([1,1]));
deleteDuplicates(getListFromArray([1,2,2,3,3]));
本地运行结果
list: 1->2->5->end
list:5->end
list: end
list:1->end
是不是很方便！
矩阵(二维数组)
矩阵的题目也有很多，基本每一个需要用到二维数组的题，都涉及到初始化，求行数列数，遍历的代码。于是简单提取出来几个函数。
/
初始化一个二维数组
@param {number} r 行数
@param {number} c 列数
@param {} init 初始值/const initMatrix= (r, c, init = 0) => new Array(r)fill()map(_ => newArray(c)fill(init));/
获取一个二维数组的行数和列数
@param {any[][]} matrix
@return [row, col]/const getMatrixRowAndCol= (matrix) => matrixlength === 0 [0, 0] : [matrixlength, matrix[0]length];/
遍历一个二维数组
@param {any[][]} matrix
@param {Function} func/const matrixFor= (matrix, func) =>{
matrixforEach((row, i)=>{
rowforEach((item, j)=>{
func(item, i, j, row, matrix);
});
})
}/
获取矩阵第index个元素 从0开始
@param {any[][]} matrix
@param {number} index/
functiongetMatrix(matrix, index) {
let col= matrix[0]length;
let i= ~~(index /col);
let j= index - i col;returnmatrix[i][j];
}/
设置矩阵第index个元素 从0开始
@param {any[][]} matrix
@param {number} index/
functionsetMatrix(matrix, index, value) {
let col= matrix[0]length;
let i= ~~(index /col);
let j= index - i col;return matrix[i][j] =value;
}
找一个简单的矩阵的题示范一下用法。@leetcode 566。题意就是将一个矩阵重新排列为r行c列。
/ @lc app=leetcode id=566 lang=javascript

[566] Reshape the Matrix/
/
@param {number[][]} nums
@param {number} r
@param {number} c
@return {number[][]}/
var matrixReshape = function(nums, r, c) {//将一个矩阵重新排列为r行c列
//首先获取原来的行数和列数
let [r1, c1] =getMatrixRowAndCol(nums);
log(r1, c1);//不合法的话就返回原矩阵
if (!r1 || r1 c1 !== r c) returnnums;//初始化新矩阵
let matrix =initMatrix(r, c);//遍历原矩阵生成新矩阵
matrixFor(nums, (val, i, j) =>{
let index= i c1 + j; //计算是第几个元素
log(index);
setMatrix(matrix, index, val);//在新矩阵的对应位置赋值
});returnmatrix;
};
let x= matrixReshape([[1],[2],[3],[4]], 2, 2);
log(x)
二叉树
当我做到二叉树相关的题目，我发现，我错怪链表了，呜呜呜这个更恶心。
当然对于二叉树，只要你掌握先序遍历，后序遍历，中序遍历，层序遍历，递归以及非递归版，先序中序求二叉树，先序后序求二叉树，基本就能AC大部分二叉树的题目了(我瞎说的)。
二叉树的题目 input 一般都是层序遍历的数组，所以写了层序遍历数组和二叉树的转换，方便调试。
function TreeNode(val, left = null, right = null) {thisval =val;thisleft =left;thisright =right;
}/
通过一个层次遍历的数组生成一棵二叉树
@param {any[]} array
@return {TreeNode}/
functiongetTreeFromLayerOrderArray(array) {
let n=arraylength;if (!n) return null;
let index= 0;
let root= new TreeNode(array[index++]);
let queue=[root];while(index
let top=queueshift();
let v= array[index++];
topleft= v == null null : newTreeNode(v);if (index
let v= array[index++];
topright= v == null null : newTreeNode(v);
}if(topleft) queuepush(topleft);if(topright) queuepush(topright);
}returnroot;
}/
层序遍历一棵二叉树 生成一个数组
@param {TreeNode} root
@return {any[]}/
functiongetLayerOrderArrayFromTree(root) {
let res=[];
let que=[root];while(quelength) {
let len=quelength;for (let i = 0; i < len; i++) {
let cur=queshift();if(cur) {
respush(curval);
quepush(curleft, curright);
}else{
respush(null);
}
}
}while (reslength > 1 && res[reslength - 1] == null) respop(); //删掉结尾的 null
returnres;
}
一个例子，@leetcode 110，判断一棵二叉树是不是平衡二叉树。
/
@param {TreeNode} root
@return {boolean}/
var isBalanced = function(root) {if (!root) return true; //认为空指针也是平衡树吧
//获取一个二叉树的深度
const d = (root) =>{if (!root) return 0;return _max(d(rootleft), d(rootright)) + 1;
}
let leftDepth=d(rootleft);
let rightDepth=d(rootright);//深度差不超过 1 且子树都是平衡树
if (_min(leftDepth, rightDepth) + 1 >=_max(leftDepth, rightDepth)&& isBalanced(rootleft) && isBalanced(rootright)) return true;return false;
};
log(isBalanced(getTreeFromLayerOrderArray([3,9,20,null,null,15,7])));
log(isBalanced(getTreeFromLayerOrderArray([1,2,2,3,3,null,null,4,4])));
二分查找
参考 C++ STL 中的 lower_bound 和 upper_bound 。这两个函数真的很好用的！
/
寻找>=target的最小下标
@param {number[]} nums
@param {number} target
@return {number}/
functionlower_bound(nums, target) {
let first= 0;
let len=numslength;while (len > 0) {
let half= len >> 1;
let middle= first +half;if (nums[middle]
first= middle + 1;
len= len - half - 1;
}else{
len=half;
}
}returnfirst;
}/
寻找>target的最小下标
@param {number[]} nums
@param {number} target
@return {number}/
functionupper_bound(nums, target) {
let first= 0;
let len=numslength;while (len > 0) {
let half= len >> 1;
let middle= first +half;if (nums[middle] >target) {
len=half;
}else{
first= middle + 1;
len= len - half - 1;
}
}returnfirst;
}
照例，举个例子，@leetcode 34。题意是给一个排好序的数组和一个目标数字，求数组中等于目标数字的元素最小下标和最大下标。不存在就返回 -1。
/ @lc app=leetcode id=34 lang=javascript

[34] Find First and Last Position of Element in Sorted Array/
/
@param {number[]} nums
@param {number} target
@return {number[]}/
var searchRange = function(nums, target) {
let lower=lower_bound(nums, target);
let upper=upper_bound(nums, target);
let size=numslength;//不存在返回 [-1, -1]
if (lower >= size || nums[lower] !== target) return [-1, -1];return [lower, upper - 1];
};
在 VS Code 中刷 LeetCode
前面说的那些模板，难道每一次打开新的一道题都要复制一遍么？当然不用啦。
首先配置代码片段 选择 Code -> Preferences -> User Snippets ，然后选择 JavaScript
然后把文件替换为下面的代码：
{"leetcode template": {"prefix": "@lc","body": ["const _max = Mathmaxbind(Math);","const _min = Mathminbind(Math);","const _pow = Mathpowbind(Math);","const _floor = Mathfloorbind(Math);","const _round = Mathroundbind(Math);","const _ceil = Mathceilbind(Math);","const log = consolelogbind(console);","// const log = _ => {}","/ 链表 /","/"," 链表节点"," @param {} val"," @param {ListNode} next"," /","function ListNode(val, next = null) {"," thisval = val;"," thisnext = next;","}","/"," 将一个数组转为链表"," @param {array} array"," @return {ListNode}"," /","const getListFromArray = (array) => {"," let dummy = new ListNode()"," let pre = dummy;"," arrayforEach(x => pre = prenext = new ListNode(x));"," return dummynext;","}","/"," 将一个链表转为数组"," @param {ListNode} list"," @return {array}"," /","const getArrayFromList = (list) => {"," let a = [];"," while (list) {"," apush(listval);"," list = listnext;"," }"," return a;","}","/"," 打印一个链表"," @param {ListNode} list "," /","const logList = (list) => {"," let str = 'list: ';"," while (list) {"," str += listval + '->';"," list = listnext;"," }"," str += 'end';"," log(str);","}","/ 矩阵(二维数组) /","/"," 初始化一个二维数组"," @param {number} r 行数"," @param {number} c 列数"," @param {} init 初始值"," /","const initMatrix = (r, c, init = 0) => new Array(r)fill()map(_ => new Array(c)fill(init));","/"," 获取一个二维数组的行数和列数"," @param {any[][]} matrix"," @return [row, col]"," /","const getMatrixRowAndCol = (matrix) => matrixlength === 0 [0, 0] : [matrixlength, matrix[0]length];","/"," 遍历一个二维数组"," @param {any[][]} matrix "," @param {Function} func "," /","const matrixFor = (matrix, func) => {"," matrixforEach((row, i) => {"," rowforEach((item, j) => {"," func(item, i, j, row, matrix);"," });"," })","}","/"," 获取矩阵第index个元素 从0开始"," @param {any[][]} matrix "," @param {number} index "," /","function getMatrix(matrix, index) {"," let col = matrix[0]length;"," let i = ~~(index / col);"," let j = index - i col;"," return matrix[i][j];","}","/"," 设置矩阵第index个元素 从0开始"," @param {any[][]} matrix "," @param {number} index "," /","function setMatrix(matrix, index, value) {"," let col = matrix[0]length;"," let i = ~~(index / col);"," let j = index - i col;"," return matrix[i][j] = value;","}","/ 二叉树 /","/"," 二叉树节点"," @param {} val"," @param {TreeNode} left"," @param {TreeNode} right"," /","function TreeNode(val, left = null, right = null) {"," thisval = val;"," thisleft = left;"," thisright = right;","}","/"," 通过一个层次遍历的数组生成一棵二叉树"," @param {any[]} array"," @return {TreeNode}"," /","function getTreeFromLayerOrderArray(array) {"," let n = arraylength;"," if (!n) return null;"," let index = 0;"," let root = new TreeNode(array[index++]);"," let queue = [root];"," while(index < n) {"," let top = queueshift();"," let v = array[index++];"," topleft = v == null null : new TreeNode(v);"," if (index < n) {"," let v = array[index++];"," topright = v == null null : new TreeNode(v);"," }"," if (topleft) queuepush(topleft);"," if (topright) queuepush(topright);"," }"," return root;","}","/"," 层序遍历一棵二叉树 生成一个数组"," @param {TreeNode} root "," @return {any[]}"," /","function getLayerOrderArrayFromTree(root) {"," let res = [];"," let que = [root];"," while (quelength) {"," let len = quelength;"," for (let i = 0; i < len; i++) {"," let cur = queshift();"," if (cur) {"," respush(curval);"," quepush(curleft, curright);"," } else {"," respush(null);"," }"," }"," }"," while (reslength > 1 && res[reslength - 1] == null) respop(); // 删掉结尾的 null"," return res;","}","/ 二分查找 /","/"," 寻找>=target的最小下标"," @param {number[]} nums"," @param {number} target"," @return {number}"," /","function lower_bound(nums, target) {"," let first = 0;"," let len = numslength;",""," while (len > 0) {"," let half = len >> 1;"," let middle = first + half;"," if (nums[middle] < target) {"," first = middle + 1;"," len = len - half - 1;"," } else {"," len = half;"," }"," }"," return first;","}","","/"," 寻找>target的最小下标"," @param {number[]} nums"," @param {number} target"," @return {number}"," /","function upper_bound(nums, target) {"," let first = 0;"," let len = numslength;",""," while (len > 0) {"," let half = len >> 1;"," let middle = first + half;"," if (nums[middle] > target) {"," len = half;"," } else {"," first = middle + 1;"," len = len - half - 1;"," }"," }"," return first;","}","$1"],"description": "LeetCode常用代码模板"}
}

---