y*****e 发帖数: 712 | 1 这题是我面groupon的店面题,所以印象深刻。。
有两个写法,一个是5楼说的,记录前一个值,比较是不是一直在变大。我和你是一样
的写法,都是比较左右两个叶子。对于max和min两个test case,最简单的还是直接拿
出来判断一下
public boolean validate(TreeNode root){
if(root == null)
return true;
return helper(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
private boolean helper(TreeNode root, int low, int high){
if(root == null)
return true;
if(root.val == Integer.MIN_VALUE && root.left != null)
return false;
if(root.val == Integer.MAX_VALUE &... 阅读全帖 |
|
y*****e 发帖数: 712 | 2 根据前面的讨论写了个java的recursive的版本,iterative的再去想想,请大家指正
public class bt_circular_list{
private static TreeNode tail = null;
public static TreeNode convert(TreeNode root){
TreeNode head = root;
helper(root);
while(head.left != null){
head = head.left;
}
head.left = tail;
tail.right = head;
return head;
}
public static void helper(TreeNode root) {
if(root == null)
return;
helper(root.left);... 阅读全帖 |
|
b******n 发帖数: 851 | 3 smallest k, using partition
int[] topKSmallest(int[] arr, int k) {
if (k <= arr.length) return arr;
helper(arr, 0, arr.length-1, k);
return arr.copyOfRange(0, k+1);
}
void helper (int[] arr, int start, int end, int k) {
if (start>=end) return;
int partitionIdx = partition(arr, start, end);
int leftLen = partitionIdx-start+1;
if (leftLen == k) return;
else if (leftLen > k) helper(arr, start, partitionidx, k);
else helper(arr, partitionIdx+1... 阅读全帖 |
|
b**********5 发帖数: 7881 | 4 void printAllRootToLeaf(TreeNode root) {
}
void helper(TreeNode root, StringBuilder sb) {
if (root == null) return;
sb.append(root.val);
helper(root.left, sb);
helper(root.right, sb);
sb.remove(sb.length()-1);
}
=> to iterative+stack
void helper(TreeNode root) {
if (root == null) return;
while (root != null) {
sb.append(root.val);
stack.push(root);
root = root.left;
}
while (!stack.isEmpty()) {
TreeNode n = stack.pop();
... 阅读全帖 |
|
b**********5 发帖数: 7881 | 5 int findMedian (int[] A, int[] B) {
int aLen = A.length;
int bLen= B.length;
if (aLength+bLength % 2 == 0) {
return (helper(A, B, (aLen+bLen)/2, 0, aLen, 0, bLen) +
helper(A, B, (aLen+bLen)/2-1, 0, aLen, 0, bLen))/2;
}
else {
return helper(A, B, (aLen+bLen)/2, 0, aLen, 0, bLen);
}
}
int helper (int[] A, int[] B, int k, int aStart, int aEnd, int bstart, int
bEnd) {
int aLen = aEnd-aStart+1;
int bLen = bEnd-bStart+1;
if (a... 阅读全帖 |
|
g******n 发帖数: 10 | 6 Sample Input
1 2 3
2 1 3
3 2 1 5 4 6
1 3 4 2
3 4 5 1 2
Sample Output
YES
YES
YES
NO
NO
--------------------------------------
Python 版:
import sys
def check(nodes):
return helper(nodes, 0, len(nodes), -sys.maxsize-1, sys.maxsize)
def helper(nodes, start, end, min_value, max_value):
if start >= end:
return True
root = nodes[start]
if not (min_value < root < max_value):
return False
i = start + 1
while i < end and node... 阅读全帖 |
|
b**w 发帖数: 78 | 7 Python的,写的比较乱
class Solution(object):
def addOperators(self, num, target):
"""
:type num: str
:type target: int
:rtype: List[str]
"""
result = self.helper(num, 0, target)
return result
def helper(self, num, pos, target):
result = []
if pos>=len(num):
if target==0:
result.append("")
return result
if pos==len(num)-1:
if int(num[pos])==target:
... 阅读全帖 |
|
|
y*******d 发帖数: 1674 | 9 LC 39
code 如下
两处不明白
1. 为什么要先排序?
2. 为什么helper(res, target-candidates[i], tmp, candidates, i); 是i不是
i+1??
多谢指点
public List> combinationSum(int[] candidates, int target) {
List> res = new ArrayList>();
if (candidates == null || candidates.length == 0) {
return res;
}
Arrays.sort(candidates);
helper(res, target, new ArrayList(), candidates, 0);
return res;
}
void ... 阅读全帖 |
|
w*********y 发帖数: 7895 | 10 现在接着说。
我找FAMILY CARE的时候有3个来源。
1. GOOGLE + CITY NAME + FAMILY CARE/DAY CARE
2. 加州政府的RESOURCES AND REFERRAL SERVICE
3. CRAIGSLIST
用1找到一部分FAMILY CARE后,我就开始打电话。主要关注2个问题,
一,他们是否有LICENSE,而,他们是否有2个OPENINGS。就一家家面试
去了。大约面试了4-5家。通过和他们的交流,知道了一些信息。比如
说有LICENSE也未必保证他们一定是好的FAMILY CARE。因为加州只有在
拿到LICENSE前的检查比较严格,拿到后,只有换地址或者有人举报时候才
会来查了。但是如果加入了加州的什么NUTRITION PROGRAM,因为加州
政府要倒贴钱给他们,所以他们一年会来几次抽查。这样更能保证他们的
服务。有一家说,没有人会不愿意拿免费的钱,除非那家FAMILY CARE
有猫腻。(当然,她的话有些偏执,因为我后面面试的一家说有些家长
不太喜欢政府的人,大意是说政府的人也会去骚扰家长问卷调查啥的。
但我们觉的知道这个... 阅读全帖 |
|
m**1 发帖数: 888 | 11 我知道这些情况是因为那个helper的女儿在我同事的丈夫手下工作。helper向她的女儿
抱怨每天太累,说owner只知道照顾自己的小孩。然后她女儿和我同事的丈夫在一起吃
午饭的时候无意中说起的。然后我同事的丈夫回家后告诉了我同事,我同事在闲聊的时
候告诉了我,然后我问了那个helper工作的地点和名字,没想到就是我儿子去的这家。
有好几次我早接我儿子,敲门站在外面等了半天,才听到owner从楼上跑下来。她家的
楼梯就在大门旁边,所以上下楼的声音都能听见。而且每次drop 和pick up都是owner
来开门,从来没有见过helper来开过门。 |
|
B****e 发帖数: 124 | 12 有人发信问小时工的详细情况,把我的回复帖出来和大家分享。我自己也还在摸索中。
也想听听大家的经验。
××××××××××××××××××××××××××××××××
我是每周4到5次(周五有时晚上出去),周末不请,因为常常出去。
每天3个小时,12块一小时。我知道有人付10块,也有人只请2小时的,但那样恐怕应征
的人不会特别多。
我家的活儿包括,洗水池里的碗,做饭(米饭是三菜一汤/粥,现在这个helper是北方
人,也时常做面食),打扫卫生,比如每天清扫厨房,擦地,每周两次清洁卫生间,每
周一次吸尘。叠烘干机里的衣服。一般也就这样了。
当然,请人总会有不合自己意的地方,换来换去也费时费力。我找过很多,目前的经验
来分享一下:
我这个地方华人多,会接到很多应征电话,我一般不一个个接电话,请他们留言自我介
绍,然后挑听起来靠谱,说话条理清晰的回电话
职责按条写下来,第一次见面逐条解释,然后帖冰箱上
当然要和helper搞好关系,但不要一见面就让人觉得你很好说话,没有什么要求,上来
就和人家掏心掏肺那种。尽量professional一点。
自己最在乎的,比如,生熟分开,倒完垃圾一定要洗手再做... 阅读全帖 |
|
z********6 发帖数: 852 | 13 古兰经有164处关于对异教徒发动圣战(jihad)的描述
The Koran's 164 Jihad Verses
K 2:178-179
Set 1, Count 1+2 [2.178] O you who believe! retaliation is prescribed for
you in the matter of the slain, the free for the free, and the slave for
the slave, and the female for the female, but if any remission is made to
any one by his (aggrieved) brother, then prosecution (for the bloodwit)
should be made according to usage, and payment should be made to him in a
good manner; this is an alleviation from your Lord and a mer... 阅读全帖 |
|
|
i***r 发帖数: 1035 | 15 对不起标题党一次~
ebay上买了个factory refurb的iphone,寄过来一个fair condition的used iphone
联系seller,他说不可能,要我发照片,说东西是她helper寄得,要找他helper核实。
发了照片,过两天他还是说不可能,说他的helper sware 我发的不是他寄得。然后
seller提出要我自付邮费,退回去她检查后退钱给我。
我于是去开了个claim,现在seller 咬定说我换了他的iphone, 说他有挤出来iphone
的照片,SN之类,我的照片和他的不符。
在ebay上回复斗嘴挺烦的,目前还没有去回复,不知道ebay怎么处理,请问大佬们此事
如何对付?
不想跟seller斗嘴,耗时间
多谢了 |
|
S**C 发帖数: 2964 | 16
it is a leveraged closed-end fund disguised as a holding company.
Is SEQUX an actively-managed fund or passively-managed fund? It is a novel
idea that active managers must churn his portfolio, or his fund falls into
passive investment/funds, aka index funds, category. It is probably defined
in such a way in another planet, but not on our Earth. Many active funds
turnover ratio is low, including SEQUX. So? That did not make them passive
funds, these folks are quite active when big opportunities ... 阅读全帖 |
|
k****n 发帖数: 369 | 17 do it directly as you do by hand?
int division(int numerator, int denominator) {
HashMap remainders = new HashMap();
StringBuffer sb = new StringBuffer();
sb.append(numerator / denominator).append(".");
return helper(numerator % denominator, denominator, remainders, sb);
}
int helper(int n, int d, HashMap remainders, StringBuffer
result) {
if (n == 0) return 0; // non-recurring
if (remainders.containsKey(n)) {
return result.toStr... 阅读全帖 |
|
s******o 发帖数: 2233 | 18 是不是正解不知道,不过pass了OJ的large case:
bool validate(int col, const vector& cols) {
int row = cols.size();
for (int i = 0; i < row; ++i) {
if (col == cols[i] || (row - i == abs(col - cols[i]))) {
return false;
}
}
return true;
}
int Helper(int n, vector& cols) {
if (cols.size() == n) {
return 1;
}
int sum = 0;
for (int i = 0; i < n; ++i) {
if (validate(i, cols)) {
cols.push_back(i);
sum += Helper(n, cols);
cols.pop_back();
}
}
ret... 阅读全帖 |
|
h****n 发帖数: 1093 | 19 来自主题: JobHunting版 - 贡献一道题 写了一个,没测,楼主测测有问题告诉我
bool isWord(string s)
{
if(s=="this") return true;
if(s=="is") return true;
if(s=="desk") return true;
if(s=="top") return true;
if(s=="desktop") return true;
return false;
}
void Helper(string &inputS, int start, string &tmpStr, vector &res)
{
if(start == inputS.size())
{
//删除多余的空格
tmpStr.erase(tmpStr.end()-1);
res.push_back(tmpStr);
//补回来以便正确的backtracking
tmpStr.push_back(' ');
return;
}
int i;
stri... 阅读全帖 |
|
r**d 发帖数: 90 | 20 来自主题: JobHunting版 - 贡献一道题 using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Collections;
namespace ConsoleApplication1
{
class Program
{
static private bool isWord(string s)
{
switch (s)
{
case "this":
case "is":
case "a":
case "all":
case "some":
case "allsome":
return true;
}
return fal... 阅读全帖 |
|
g***j 发帖数: 1275 | 21 一下都是我面试的经验和教训,欢迎各位大牛指正或者补充
首先,要端正观念,写代码只是最后一步,是在对方完全理解了你的意图之后的最终表
述,所以,在写代码之前,一定要跟对方把你的意图表述清楚,一定不要在对方不懂你
的想法的情况下就开始写代码,大忌,大忌!
其次,写代码之前,大脑里面要有个大picture,不能想到哪儿写到哪儿。是你的大脑
在写代码,而不是白板上你的手在代码。你的手只是一个printer里面的喷头而已,是
它把你大脑里面的代码print到白板上,你的大脑才是控制那个喷头的芯片。所以,写
之前,你要看着那个白板打个腹稿,想想一下白板上可能有哪些代码,比如定义哪些变
量,哪些if else,哪里退出,call哪几个function,等等。
第三,你在白板或者纸上写代码的过程中,一定要跟面试官交流,让他知道自己在干什
么。每次提笔之前,告诉他,我前面写了啥,然后我准备写啥,这个写的过程,是前面
跟面试官讨论问题结束之后的具体反映。
第四,如果有重复的代码,一定要用一个变量或者一个function表示。本来面试的代码
就不长,还有重复的代码会很ugly。比如类似current->ne... 阅读全帖 |
|
p*****2 发帖数: 21240 | 22 我也写了一个练练。
mat=[[1,2,3,4],[5,6,7,8],[9,10,11,12]]
def helper(mat, i, j)
while i=0
print "#{mat[i][j]} "
i+=1
j-=1
end
puts
end
def print_mat(mat)
(0...mat[0].length).each {|k| helper(mat,0,k)}
(1...mat.length).each {|k| helper(mat,k,mat[0].length-1)}
end
print_mat(mat) |
|
j*****y 发帖数: 1071 | 23 写了一个,欢迎测试
#include
#include
#include
#include
#include
using namespace std;
bool isNumber(string s)
{
if(s[0] == '+' || s[0] == '-' || s[0] == '(' || s[0] == ')')
{
return false;
}
return true;
}
int helper(int a, int b, char c)
{
if(c == '+')
{
return a + b;
}
else
{
return a - b;
}
}
bool valid(vector &s)
{
stack v;
stack c;
for(int i = 0; i < s.size(); ++i)
... 阅读全帖 |
|
j*****y 发帖数: 1071 | 24 写了一个,欢迎测试
#include
#include
#include
#include
#include
using namespace std;
bool isNumber(string s)
{
if(s[0] == '+' || s[0] == '-' || s[0] == '(' || s[0] == ')')
{
return false;
}
return true;
}
int helper(int a, int b, char c)
{
if(c == '+')
{
return a + b;
}
else
{
return a - b;
}
}
bool valid(vector &s)
{
stack v;
stack c;
for(int i = 0; i < s.size(); ++i)
... 阅读全帖 |
|
z***y 发帖数: 50 | 25 void rev(vector &arr, int i, int j){
while(i < j){
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
i++;
j--;
}
return;
}
void Helper(vector &arr, int left, int right){
if(left == right) return;
int mid = left + (right-left)/2;
Helper(arr, left, mid);
Helper(arr, mid+1, right);
int i = mid, j = mid+1;
if(arr[i] < 0 || arr[j] > 0) return;
if(arr[i] <= 0 && arr[j] >= 0) return;
for(; i >= left && arr[i] >= 0;... 阅读全帖 |
|
b*********s 发帖数: 115 | 26 def solution(n):
cache = {}
def helper(target, minVal, allowedZero):
if (target, minVal) in cache:
return cache[(target, minVal)]
if minVal > target:
return []
elif minVal == target:
return [[minVal]]
else:
res = []
for i in range(minVal, target):
tail = helper(target - i, i, True)
for solution in tail:
solution.append(i)
res += t... 阅读全帖 |
|
C*******n 发帖数: 24 | 27 标记个start,然后一步一步开始做
java code,貌似我童鞋也被考这个题了。
个人做了下,蛮有思路,耗时有点长,但是差不多能handle
不过真的算是个难的phone题啊,我这个小弱是这么觉得的。。。。
public void solution(int num){
int i = 1;
while(i * i <= num){
if(i == 1){
System.out.println(num + " * 1");
}else if( num % i == 0){
ArrayList> tmp = helper(num / i, i);
for (ArrayList r : tmp){
for ( Integer no : r){
System.out.print(no + " * ");
}
System.out.println(i);
... 阅读全帖 |
|
w********8 发帖数: 55 | 28 借鉴楼上各位大牛的做法,我也写了一份Java的代码。。。。
public void mySolution(int num) {
if (num >= 5) {
String str = String.valueOf(num);
helper(num, "", 0, str.length(), false);
}
}
private void helper(int max, String prefix, int pos, int len, boolean
have5) {
if (pos == len) {
int cur = Integer.parseInt(prefix);
if (cur <= max) {
System.out.print(cur + ", ");
}
return;
}
for (int... 阅读全帖 |
|
w********8 发帖数: 55 | 29 借鉴楼上各位大牛的做法,我也写了一份Java的代码。。。。
public void mySolution(int num) {
if (num >= 5) {
String str = String.valueOf(num);
helper(num, "", 0, str.length(), false);
}
}
private void helper(int max, String prefix, int pos, int len, boolean
have5) {
if (pos == len) {
int cur = Integer.parseInt(prefix);
if (cur <= max) {
System.out.print(cur + ", ");
}
return;
}
for (int... 阅读全帖 |
|
f**********e 发帖数: 288 | 30 Determine whether an integer is a palindrome. Do this without extra space.
public class Solution {
public boolean isPalindrome(int x) {
if(x < 0)
return false;
int y = x;
return helper(x, y);
}
public boolean helper(int x, int y){
if(x == 0)
return true;
if (helper(x/10, y) && (x % 10 == y % 10) ){
y =y/ 10;
return true;
}
else
return false;
... 阅读全帖 |
|
d****n 发帖数: 233 | 31 Given n distinct positive integers, integer k (k <= n) and a number target.
Find k numbers where sum is target. Calculate how many solutions there are?
Example
Given [1,2,3,4], k=2, target=5. There are 2 solutions:
[1,4] and [2,3], return 2.
我用递归如下:
void helper(vector A,int k, int start,int target, int & ans) {
if (k < 0 || target < 0) return;
if (k == 0 && target == 0) {
ans++;
return;
}
for(int i = start; i <= A.size(... 阅读全帖 |
|
a***e 发帖数: 413 | 32 https://oj.leetcode.com/problems/distinct-subsequences/
DP的题在现场除了写出来,还要求优化space吗?像这个要改成一维的table吗?
class Solution {
public:
int numDistinct(string S, string T) {
int slen=S.length(), tlen=T.length();
if (slen
vector> t(slen+1,vector((tlen+1), 0));
for(int i=0; i<=slen; i++)
t[i][0]=1;
for (int i=1; i<=slen; i++)
for (int j=1; j<=tlen; j++)
{
if (S... 阅读全帖 |
|
b******i 发帖数: 914 | 33 Hi, not so hard using recursion:
Explanation:
For every index, I maintain three states: ONEBYTE (single byte),
TWOBYTEFIRST (1st element of a two-byte), TWOBYTELAST (2nd element of a two-
byte). If you now the state of i-1'th element, you can deduce the state of i
'th.
One minor case is that if the last element is identified as the first byte
of a two-byte, the below logic still returns one-byte. But this can be
modified.
------------------------
class Solution {
public:
int last_byte_class(... 阅读全帖 |
|
x*******6 发帖数: 262 | 34 贡献一个code,由于没有乘除,只需要记录括号内是否要根据加减改变数字的符号,不
需要使用reverse polish notation解法。
public static int eval(String s, int x) {
String[] exp = s.split("=");
int[] left = helper(exp[0],x);
int[] right = helper(exp[1],x);
return (left[0] - right[0])/(right[1]-right[0]);
}
private static int[] helper(String s, int x) {
boolean positive = true;
Stack re = new Stack();
re.push(false);
int num = 0;
int y = 0;
... 阅读全帖 |
|
C****t 发帖数: 53 | 35 def revSum(a):
d = height(a)
level, sum = 1, [0]
helper(a, d, level, sum)
return sum[0]
def height(a):
d = 1
for ele in a:
if type(ele) == list:
d = max(d, 1+height(ele))
return d
def helper(a, d, level, sum):
for i in range(len(a)):
if type(a[i]) != list:
sum[0] += a[i]*(d-level+1)
else:
helper(a[i], d, level+1, sum) |
|
C****t 发帖数: 53 | 36 def revSum(a):
d = height(a)
level, sum = 1, [0]
helper(a, d, level, sum)
return sum[0]
def height(a):
d = 1
for ele in a:
if type(ele) == list:
d = max(d, 1+height(ele))
return d
def helper(a, d, level, sum):
for i in range(len(a)):
if type(a[i]) != list:
sum[0] += a[i]*(d-level+1)
else:
helper(a[i], d, level+1, sum) |
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l*****n 发帖数: 246 | 37 第2题不是应该是leetcode上的permutation II吗?还是我题意理解错了?
尝试着写了个第二题和第三题的代码:
//给定一个字典,给定一个字符串,返回所有可能的组合。
public List words(Set dict, String str) {
char[] arr = str.toCharArray();
List result = new ArrayList();
Arrays.sort(arr);
helper(dict, arr, result, new boolean[arr.length], new ArrayList<
Character>());
return result;
}
private void helper(Set dict, char[] arr, List result,
boolean[] used, List item) {
if(item.size()==arr.l... 阅读全帖 |
|
b**********5 发帖数: 7881 | 38 返回每个点左右子树的和与自己值的差?
how should we return this? in a list? or just print it out?
void helper( ) {
if (root == null) return 0;
sum1 = helper(root.left);
sum2 = helper(root.right);
print(sum1+sum2-root.val);
return sum1+sum2+root.val;
} |
|
b******n 发帖数: 851 | 39 dp为什么会没recursive好?
cur = prevPrev+{2} && prev+{1}
[[]]
[[1]]
[[1,1],[2]]
[[1,2], [1,1,1], [2, 1]]
List> EnumeratingSteps(int n) {
List> prevPrev = new ArrayList>();
if (n == 0) return prevPrev;
List> prev = new ArrayList>();
prev.add (new ArrayList().add(1));
if (n == 1) return prev;
for (int i = 2; i <= n; i++) {
List> cur = new ArrayList>();
for (List阅读全帖 |
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b******n 发帖数: 851 | 40 void helper (TreeNode node, TreeMap m, int level) {
List nodesAtThisLevel = m.get(level);
if (nodesAtThisLevel == null) ...
nodesAtThisLevel.add(node);
helper(node.left, m, level-1);
helper(node.right, m, level+1);
} |
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b******n 发帖数: 851 | 41 void helper (TreeNode node, TreeMap m, int level) {
List nodesAtThisLevel = m.get(level);
if (nodesAtThisLevel == null) ...
nodesAtThisLevel.add(node);
helper(node.left, m, level-1);
helper(node.right, m, level+1);
} |
|
b******n 发帖数: 851 | 42 post my code to see if there's any bug:
// 1 2 3 4 5 (BST)
// find 1st greatest element -> return 5
// find 3rd greatest -> return 3
TreeNode helper(TreeNode root, int[] k) {
if (root == null) return null;
TreeNode result = helper(root.right, k);
if (result != null) return result;
else {
--k[0];
if (k[0] == 0) {
return root;
}
}
return helper(root.left, k);
} |
|
e****x 发帖数: 148 | 43 这是用kth smallest element in two sorted arrays来做的吧?这样确实不用考虑
edge case了……我当时想到的是那个一般二分法的solution,aLen不等于bLen的情况
下很多edge cases的那个,不确定能写好就想先写个O(N)的算了...
: int bLen= B.length;
: if (aLength+bLength % 2 == 0) {
: return (helper(A, B, (aLen+bLen)/2, 0, aLen, 0, bLen) +
: helper(A, B, (aLen+bLen)/2-1, 0, aLen, 0, bLen))/2;
: }
: else {
: return helper(A, B, (aLen+bLen)/2, 0, aLen, 0, bLen);
: } |
|
k****r 发帖数: 807 | 44 我写了一个,不知道可用不,先用DP找到最长subsequence的长度,complexity is O(
nlongn);
再用recursive找到所有可行解。
public static List> AllLongestIncreasingSubsequence(int[] nums
) {
int[] m = new int[nums.length + 1];
int L = 0;
for (int i = 0; i < nums.length; i++) {
int lo = 1;
int hi = L;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (nums[m[mid]] < nums[i]) lo = mid + 1;
else hi = mid - 1;
}... 阅读全帖 |
|
o******y 发帖数: 446 | 45 用链表和backtrack:
public int getMaxSum(int[] arr){
LinkedList q = new LinkedList<>();
for(int n: arr) q.add(n);
int[] res = new int[1];
res[0] = Integer.MIN_VALUE;
helper(res, q, 0);
return res[0];
}
void helper(int[] res, LinkedList q, int sum){
if(q.isEmpty()){
res[0] = Math.max(sum, res[0]);
return;
}
int size = q.size();
while(size-->0){
int v = q.rem... 阅读全帖 |
|
l***c 发帖数: 1634 | 46 他们会review code吗?还是直接看成绩?
90分钟两道题,第一道题应该在medium难度,第二道题在hard难度
其中第二道题我一共写了5个helper function,一共快100行,总是报一个npe的错,我
以为是其中一个helper function的问题,debug好久,最后还是没过
后来放到eclipse里debug了一下,原来是另外一个helper function的问题,easy fix
是不是废了?
frozen time多久? |
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m**1 发帖数: 888 | 47 我儿子送去一家home daycare. 一个owner 和一个helper. 我儿子和helper的关系很好
。很明显helper 每天和我儿子带的时间比较长。我想买份礼物送给她,不知道送什么
好,而且又怕owner 不高兴。 |
|
m******j 发帖数: 5079 | 48 From our PTA:
"Like all of us, I'm shocked and pained by today's tragic event in Newtown,
Connecticut.
As parents and PTA leaders, we may also have to help our children cope with
scary news. Below are some helpful hints from Mr. Fred Rogers, the creator
of the children's television show Mr. Rogers' Neighborhood.
Please feel free to share with your members. For more resources, you can go
to the Fred Rogers Website or to the American Psychological Association.
.....
"When I was a boy and I would ... 阅读全帖 |
|
y****i 发帖数: 5690 | 49 我发觉娃严重像现在中年的自己 在懒惰 LAID BACK方面
我小时候比娃好胜 比娃做事情有MOTIVATION很多
现在越发觉得有MOTIVATION 做事情肯努力太重要了 但是这个娃真的不是那个料子
昨儿一起游泳的娃们在小区比赛 朋友的女娃 看到水里有HELPER(怕她们小娃们游不完
25米)就不高兴了 要求不要HELPER 进水就猛游啊 我家的 进水游了4下 开始转为仰天
那样轻松嘛 然后被HELPER一路托着游完全程
我突然被刺激了 觉得非常需要调整对娃的认识和期望了 也许这就是他的天性吧 |
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r*******u 发帖数: 8732 | 50 http://www.maid.hk/web/eng/ask.jsp
How much do I need to spend to hire a Helper?
Answer4: The Agency Fees varies from the services they provided; please
refer to the Employment Agencies directly for accurate price and services.
Question 5: What is the capability for an employer to hire a helper?
Answer5: According to the Employment Ordinance, the employer must have a
household income of HK$178,000 per annually to hire one foreign domestic
helper.
Question 6: Can I hire a |
|