t****3
发帖数: 2337 | 1 Quoted from vtec.net
There are quite a few difference actually...
VTM-4:
-Front wheel drive under normal driving
-Can transfer power to rear differential when needed, but power distribution
is limited to front to rear and not side to side.
-Sends power to rear when needed (ie limited traction)
-Sends power to rear when accelerating at low speeds up to 18mph.
-Sends power to rear when in reverse
-Sends power to rear when diff lock is on and tranny is in D1 or D2 up to
18mph.
SH-AWD: Based on VTM-... 阅读全帖 |
|
t****3
发帖数: 2337 | 2 Quoted from vtec.net
There are quite a few difference actually...
VTM-4:
-Front wheel drive under normal driving
-Can transfer power to rear differential when needed, but power distribution
is limited to front to rear and not side to side.
-Sends power to rear when needed (ie limited traction)
-Sends power to rear when accelerating at low speeds up to 18mph.
-Sends power to rear when in reverse
-Sends power to rear when diff lock is on and tranny is in D1 or D2 up to
18mph.
SH-AWD: Based on VTM-... 阅读全帖 |
|
t****3
发帖数: 2337 | 3 我来说说
"那为什么用纵置引擎的Audi Quattro要学SH-AWD，在后轮上加一个用多片离合器的
sport diff呢？"
其实我说过我的观点，是成本考虑，轿车唯一有托森中差加后差的车是二十年前的audi
V8，还是选装。后来A8都取消了托森后差，因为高速上，雪地里，左右轮子扭矩差别
不大，有了电控刹车后左右差用处就更不大了，只是损耗功率和刹车片。
sport diff虽然用到多片离合器，但不是学SH-AWD，SH-AWD的左右后轮100%的扭矩是靠
多片离合器，而sport diff的设计是正常行驶时左右是50:50的齿轮咬合传输，多片离
合器不工作，可以传输更大的扭矩，完成真正的扭矩均衡分配的全时4驱。当一后轮需
要大扭矩时，用多片离合器将更多的扭矩叠加在齿轮传输的主扭矩上，而且是电控液压
泵加压，压力大，电流小，发热小。高速转弯时再配合电控刹车，就可以使外后轮持续
得到更多扭矩，更高转速。 |
|
h*****g
发帖数: 357 | 4 我不知道有的车是front diff 和trans 整合在一起. 或许有吧.
这两工作环境不同, 对油品的要求也不同. 我换油的时候, trans fluid和transfer
case fluid都是Nissan Matic D ATF.
Font Diff我用的是 Mobil1 Synthetic Gear Lubricant 75W-90. 更冷的地方可以用
80W-90.
因为是part time四驱. Front Diff 大部分时候都是歇着. 全时四驱可能是不同的工作
环境. |
|
m***j
发帖数: 533 | 5 trans和diff是两个截然不同的story。trans fluid有清洁trans的duty，用时间长了和
机油一样会变脏（机油其实需要更换的原因也是变脏），而且filter会clog，tranny
fluid工作温度在140C所以会变质，所以理论上都是需要定期更换的。diff里面不存在
清洁的问题，diff oil唯一的duty就是润滑齿轮，工作温度也基本不会超过80C（tow除
外），基本上是不需要换的。 |
|
m******6
发帖数: 210 | 6 Drove H3 on icy road for a few weeks in Rock mountain area. As you know, H3
can turn on/off central diff. If it is in the open mode, car swings a little
on black ice. If I turn on central diff, it is hard to tell that there is a
black ice.
Drove Tahoe 4WD in Quebec for two weeks in winter break. I can turn on/off
4WD on Tahoe. The car was out of control under 20 mph in 2WD mode (try to
save gas). Since then, I switched to 4WD mode all the time. and it never
happened again.
Haven't tried MDX on i... 阅读全帖 |
|
a********c
发帖数: 3657 | 7 你说话跟老色狼一个调调，都是自以为是，那你说说为啥quattro起步爬坡可能比电子
鲜花的差吧？
机械diff只可能在一种情况下差，就是tire完全没有grip，diff不知道怎么怎么分配
torque，但是这种情况现实中基本不存在。即使发生机械diff的车还有trac，可以通过
abs修正
0 grip的问题。
xdrive跟haldex就是一个东西，区别是xdrive常态后驱，hadex常态前驱。德国鬼子的
auto motor sports专门对比过xdrive和haldex的雪地表现，结果是haldex好些。
懂德语可以去收收2013的后面几期。 |
|
e******g
发帖数: 5344 | 8 这一代野马不了解，但是上一代野马质量很差，开到8万迈的时候就到处漏了，我上一
辆漏的地方有：
变速箱油，机油，刹车液，还有washer fluid tank和rear diff。
变速箱油和机油除了把地搞脏问题不大，因为漏得很慢，刹车液经常加也还好，rear
diff开着开着突然烂了，拖去修才发现是油漏光了，当然只能换了个新的diff，花了
2000刀。
washer fluid的问题就是一加就漏光，所以换了个tank，其实就是gasket漏了，tank本
身还是好的，但是连体设计，只能整个换, 这个花了200刀
另外还有washer fluid的喷嘴一边不出水，换了个motor就好了，这个花了80刀。
还有年岁大了，右前轮的caliper也坏了，换了个新的，花了400多刀。
然后我就受不了了。卖了。
所以我要是再买野马肯定lease，新的ecoboost由于是turbo，tune的潜力肯定很大，买
辆刷一下ECU改一下排气和intake操3年还了肯定不错。 |
|
t**********r
发帖数: 182 | 9 ERROR: The following columns were not found in the contributing tables: diff.
516 having diff=min(diff);
------
180
ERROR 180-322: Statement is not valid or it is used out of proper order. |
|
t**********r
发帖数: 182 | 10 Many thanks for your hint!! I made it. Here is the code.
579 proc sql;
580 create table data3 as
581 select data1.*, data2.rating, day1-day2 as diff
582 from data1, data2
583 where data1.var1=data2.var1 and data1.var2=data2.var2
584 and date1-date2>0
585 group by data1.var1, data1.var2, data1.date1
586 having diff=min(diff);
NOTE: The query requires remerging summary statistics back with the original
data.
NOTE: Table WORK.data3 created, with 48144 rows and 9 |
|
h*****n
发帖数: 924 | 11 What is cost diff of education between A and B? Diff in their tuitions?
Anyway, you realize the language problem. Good.
Now let's move on to the logic problem.
What you said implies that high type is more likely to receive education tha
n low type. If your "cost diff" is larger, education is more likely used
as a signal. This makes sense, but off the topic.
Remember that lz's question is why Chinese care more about ranking, i.e.,
value high-quality education more.
I hope you get it now. Try agai |
|
r****o
发帖数: 1950 | 12 明白了，
那这样行吗？
用b-a的二进制形式，比如说是12=1100，共4位。
那么运行random(0,1)4次，每次相当于二进制的一位，如果凑成的二进制数小于等于
1100，则返回randNum+b-a，否则重来。
int diffSize=0;
int diff=b-a;
while(!diff)
{
diffSize++;
diff>>=1;
}
int randNum=MAX_INT;
while(randNum>b-a)
{
randNum=0;
for (int i=0; i
{
randNum+=random(0,1)<
}
}
return randNum+b-a;
learn
learn |
|
s****n
发帖数: 36 | 13 刚刚结束second phone interview，求祝福
第一次
1. diff of class of object
2. diff of overload and overriding
3. diff of array and linked list
4. How to test a web page with a search box
5. coding: find integer appear odd times in an array.
following: when more than one integers appear odd times, how to handle?
How to test?
第二次
1. Previous experience
2. How to find the second largest integer in an array
How to test?
3. OOD for CD Player |
|
j*****u
发帖数: 1133 | 14 1. 我一开始想成根据history的stock price来做predict了，想了半天。。。
其实是这个题的变种
int array A里找两个position i和j，要求A[j]和A[i]的差maximum，同时i
static void BestDay(int[] stockPrices, out int buyDate, out int sellDate)
{
buyDate = sellDate = 0;
int maxDiff = 0, minPrice = stockPrices[0], minPriceDate = 0;
for (int i = 1; i < stockPrices.Length; ++i)
{
int price = stockPrices[i];
int diff = price - minPrice;
if (diff > maxDiff)
{
buyDate = minPriceDate;
s... 阅读全帖 |
|
s********e
发帖数: 1875 | 15 CEO will also limit your options a lot for job switch, but would you like to
be CEO? :-)
The issue here is:
if( Diff = CEO - SDE)
take CEO
else if(Diff = scientist - SDE)
evaluate the significance of Diff, which is difficult since I don't know. |
|
j*****u
发帖数: 1133 | 16 bool F(string s, char ch1, char ch2, int n)
{
if (string.IsNullOrEmpty(s) || n <= 0 || ch1 == ch2)
throw new ArgumentException();
var index1 = new List();
var index2 = new List();
for (int i = 0; i < s.Length; i++)
{
if (s[i] == ch1)
index1.Add(i);
else if (s[i] == ch2)
index2.Add(i);
}
int minDist = int.MaxValue;
int i1 = 0, i2 = 0;
while (i1 < index1.Count && i2 < index2.Count)
{
int diff ... 阅读全帖 |
|
j*****u
发帖数: 1133 | 17 bool F(string s, char ch1, char ch2, int n)
{
if (string.IsNullOrEmpty(s) || n <= 0 || ch1 == ch2)
throw new ArgumentException();
var index1 = new List();
var index2 = new List();
for (int i = 0; i < s.Length; i++)
{
if (s[i] == ch1)
index1.Add(i);
else if (s[i] == ch2)
index2.Add(i);
}
int minDist = int.MaxValue;
int i1 = 0, i2 = 0;
while (i1 < index1.Count && i2 < index2.Count)
{
int diff ... 阅读全帖 |
|
k*j
发帖数: 153 | 18 i think i was asked the same question long time ago when i interviewed with
them
my answer was like:
for two searches, you get something like the following, each is a link, and
they are in order.
search 1: ABCDEF
search 2 : FABEDG
and for each link in the search1, you find its position in search2, and then
calculate the absolute difference of two positions. for example, for link "A", it is
position 0 in search 1, and position 1 in search2. the diff is abs(1-0)=1.
another example, for link "F", i... 阅读全帖 |
|
s******n
发帖数: 3946 | 19 第二题,搞个升序的dequeue, dequeue[0]表示当前k个连续元素最小difference。
每来一个元素diff[i]，把dequeue上所有比diff[i]大的扔掉，如果dequeue[0]=diff[i
-k]则把dequeue[0]也扔掉。 |
|
w****x
发帖数: 2483 | 20
这题他说的没错, 就是按search的路径找diff最小的, 你可以证明
如果当前节点比目标数字小, 你只需要search右子树, 因为左子树所有节点的diff肯定
更大,
如果当前节点比目标数字大, 你只需要search左子树, 因为右子树所有节点的diff肯定
更大,
相等的话那就是最小喽.
把上面3条想通了就可以了, 我想了7,8分钟, NND |
|
g****y
发帖数: 240 | 21 来自主题: JobHunting版 - 请问G一题 dynamic programming solution.
input: array, with all elements greater than or equal to zero.
p[i,j,k]: Ture, if for array[0..i], there exists a subset of length k, whose
sum equals to j; False otherwise
optimal substructure: p[i,j,k] = True if p[i-1,j,k] is True or p[i-1, j-
array[i], k -1] is True
After calculating p, the original problem becomes: for i = len(array)- 1, k
= len(array) / 2, find j such that p[i,j,k]=True and abs(sum - 2 * j) is
minimized.
def balanced_partition_even(array):
... 阅读全帖 |
|
h**********9
发帖数: 3252 | 22 来自主题: JobHunting版 - 请问G一题
diff
如图可看出对角线之和是delta,找到max diff后swap X/Y, 然后重新计算黄色的cell.
上面的code是偷懒，swap X/Y 后没有重新对 X/Y 排序，否则找 max diff 就是小于 O
(NlgN)。
初始化能保证delta < max(input) - min(input), 每次while() loop使得delta递减直
到收敛，最坏是 max(input) - min(input)次，（实际递减速度应快于每次减一，具体
应该和数据的pattern有关，）现在的问题是能否证明这样每次swap一对元素使得delta递
减，最后能不能收敛到理论最小值。
谁有时间就看看这个行不行吧。 |
|
g****y
发帖数: 240 | 23 来自主题: JobHunting版 - 请问G一题 dynamic programming solution.
input: array, with all elements greater than or equal to zero.
p[i,j,k]: Ture, if for array[0..i], there exists a subset of length k, whose
sum equals to j; False otherwise
optimal substructure: p[i,j,k] = True if p[i-1,j,k] is True or p[i-1, j-
array[i], k -1] is True
After calculating p, the original problem becomes: for i = len(array)- 1, k
= len(array) / 2, find j such that p[i,j,k]=True and abs(sum - 2 * j) is
minimized.
def balanced_partition_even(array):
... 阅读全帖 |
|
h**********9
发帖数: 3252 | 24 来自主题: JobHunting版 - 请问G一题
diff
如图可看出对角线之和是delta,找到max diff后swap X/Y, 然后重新计算黄色的cell.
上面的code是偷懒，swap X/Y 后没有重新对 X/Y 排序，否则找 max diff 就是小于 O
(NlgN)。
初始化能保证delta < max(input) - min(input), 每次while() loop使得delta递减直
到收敛，最坏是 max(input) - min(input)次，（实际递减速度应快于每次减一，具体
应该和数据的pattern有关，）现在的问题是能否证明这样每次swap一对元素使得delta递
减，最后能不能收敛到理论最小值。
谁有时间就看看这个行不行吧。 |
|
d****c
发帖数: 234 | 25 CC里8.8 Write an algorithm to print all ways of arranging eight queens on a
chess board so that none of them share the same row, column or diagonal.
12行 printBoard();什么功能。是类似println()；吗？谢谢！
01 int columnForRow[] = new int [8];
02 boolean check(int row) {
03 for (int i = 0; i < row; i++) {
04 int diff = Math.abs(columnForRow[i] - columnForRow[row]);
05 if (diff == 0 || diff == row - i) return false;
06 }
07 return true;
08 }
09
10 void PlaceQueen(int row){
11 if (row == ... 阅读全帖 |
|
i********m
发帖数: 332 | 26 练习了
public ListNode reverseBetween(ListNode head, int m, int n) {
// Start typing your Java solution below
// DO NOT write main() function
int diff = n-m;
if (diff < 0 || m <= 0)
return null;
boolean flag = false;
ListNode cur = head;
ListNode prev = head;
while (m-- > 1) {
if (cur == head) {
cur = cur.next;
}
else {
cur = cur.next;
... 阅读全帖 |
|
A******g
发帖数: 612 | 27 Leetcode OJ里的word ladder, 过了小test，大test第一个就超时了，不解！
用的是BFS算法，请大牛们看看有什么问题，谢谢！
class Solution {
public:
typedef unordered_map > Graph;
bool has_edge(string a, string b) {
if (a.size() != b.size()) {
return false;
}
int diff = 0;
for (int i=0; i
if (a[i] != b[i])
diff++;
}
return diff==1;
}
void build_graph(string &start, string &end, unordered_set阅读全帖 |
|
d*******y
发帖数: 27 | 28 牛顿迭代法的时间复杂度很不好算，可以看看这个http://en.citizendium.org/wiki/Newton%27s_method#Computational_complexity
binary search应该是O(log n e)吧，n是待求数字，e是精度。不是很确定。
写了一下这两个程序：
public class FloatSqrt {
public static double error = 0.000001;
public static double getFloatSqrt(double x) {
if (x <= 0) {
return -1;
}
double y = x;
double z;
while (true) {
z = (y + x / y) / 2;
if (Math.abs(z - y) < error) {
break;
} else {
y = z;
}
}
return y;
}... 阅读全帖 |
|
h******2
发帖数: 13 | 29 有两台机器，每台10T数据， 数据中都是url，每行一个url， 他们只有万分之一的
diff， 要查找有这两台机器的url的差集， 需要一个准确的结果（不能用boolfilter)。
类似的一题是：也是两台各10T数据，一开始两边数据相同，后来可能两边有更改，如
果能够提供一个接口，快速的比较两边数据是否有diff， 如果有，diff的是哪些url。 |
|
g***9
发帖数: 159 | 30 遵照大牛的思路写了个实现：
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int getMaxEvenSeq(vector &v) {
int n = v.size();
if (n < 2) {
return 0;
}
vector pos(2*n+1, -1);
int zeros, ones;
int i, t, diff, index, dis, ans;
zeros = ones = 0;
ans = 0;
for (i = 0; i < n; i++) {
t = v[i];
if (t == 0) {
... 阅读全帖 |
|
d*******y
发帖数: 27 | 31 基本上2,3,4-sum都是一个思路吧，用hashmao做一个的索引。我之前做
的时候都是优化一个指数级别就可以过large set了。
import java.util.*;
public class Solution {
public ArrayList> fourSum(int[] num, int target) {
// Start typing your Java solution below
// DO NOT write main() function
HashSet> result = new HashSet>
();
HashMap map = new HashMap();
ArrayList> rtn = new ArrayLis... 阅读全帖 |
|
g*****g
发帖数: 212 | 32 here is the code
int maxInMatrix(vector> &matrix)
{
int n = matrix.size();
int m = matrix[0].size();
vector premin(m+1, INT_MAX);
int maxdiff = 0;
for(int i=0; i
{
vector curmin(m+1, INT_MAX);
for(int j=0; j
{
curmin[j+1] = min(premin[j+1], curmin[j], matrix[i][j]);
int diff = matrix[i][j] - curmin[j+1];
if (diff > maxdiff)
... 阅读全帖 |
|
p*****d
发帖数: 126 | 33 我感觉你这个代码由问题。
你的MAX_INT 按照字面上的意思是应该是2的31次方-1
如果value=（2的31次方-1） 而数组中的一个元素是一个负数，比如-2,那diff就等于
负的2的31次方+1，负的值小于0，而实际你想得到的是个正值（2的31次方+1）
diff应该为unsigned 这样能表示的范围会从0到2的32次方-1，如果diff是unsigned的
话value -(-2) 就是可正确的表示了，因为2的31次方+1在0到2的32次方-1内 |
|
G***n
发帖数: 877 | 34 Find three integers in S such that the sum is closest to a given number.
下面的code没问题。我想优化一下，skip那些重复的数，但加上下面这2个while语句然
后返回的结果就错了，搞不懂哪里的问题。
//while (j
// while (j
class Solution {
public:
int threeSumClosest(vector &num, int target) {
int sum = INT_MIN;
int minDist = INT_MAX;
if (num.size()<3) return sum;
sort(num.begin(), num.end());
for (int i = 0; i阅读全帖 |
|
q********c
发帖数: 1774 | 35 这是我的口德. 基本原理就这样，再讨论一下scalability, concurrency, 就差不多了。
void RateLimit() {
int minInterval = 1000/N; // N is the max rate
int current = now();
int diff = current-last;
last = current; //last would be a class private member
if(diff < minInterval) {
Thread.sleep(diff);
return ;
}
else
return ;
} |
|
v****a
发帖数: 236 | 36 LS说得对。
也可以用double 作为中间结果，就不用判断溢出了。
double diff = 0.01;
double m = (low + high) / 2;
if(Math.abs(m*m - x) < diff)
{
return (int)m;
}
else
{
... 二分查找, 注意考虑diff
} |
|
b******n
发帖数: 1629 | 37 感觉要挂。
一个server，每有人访问call log_hit(). 然后另一个函数get_log_hit_5_min()返回
近五分钟的访问数.
我傻哈哈的搞了个这个。结果要求常数时间返回，说我搞得太复杂
class hit
{
Time tag;
}
vector hit_list;
hit_log()
{
hit_list.push_back(hit);
}
int get_hit_5_min()
{
count = 0;
for (int i = hit_list.size() - 1; i >= 0; --i)
{
if (time_diff(hit[i].tag, current_time()) > 5)
count++;
else
break;
}
return count;
}
因为比如访问量很大，会有存储限制，然后改成这个，不知道还有没有更好的办法。匆
匆结束。
int count;
Time start_time;
hi... 阅读全帖 |
|
s***g
发帖数: 1250 | 38 int bugCrossRiver(vector a, int X, int D)
{
int n = a.size();
int stepToX = X;
int curPos = 0;
int prePos = 0;
for (int i=0; i
int diff = a[i]-curPos;
if (diff>0 && diff<=D) {
curPos = a[i];
stepToX -= curPos;
stepToX += prePos;
prePos = a[i];
if (stepToX<=D) {
return i;
}
}
}
return -1;
} |
|
h***b
发帖数: 1233 | 39 when uncertain, disclose it.
laws require certain disclosures but diff buyer w/ diff culture and
background may have diff concerns--no right or wrong. my Korean clients are
the most aggressive--don't care just about anything as long as it's Irvine.
weight the various aspects of the deal--price, location, time/energy spent,
alternatives (important in Irvine given the high price tag!!). lots of
opportunity costs involved here.
buying/owning a house is a long term commitment. "buy it when you f |
|
h***b
发帖数: 1233 | 40 cost of insurance is always a consideration whether it's Term or Whole.
diff ins products are designed for diff purposes. for average ppl in "
general" if you're just looking for protection to hedge against lost of
income (thus can't make house payment for example) during "peak" earning
power (e.g. age 30-55), a Term policy would probably be sufficient. again,
diff situation may consider differently. |
|
b*****e
发帖数: 53215 | 41 diff school,diff company, diff discount |
|
f*****g
发帖数: 15860 | 42 the data from HCPSS is by median score and # of participants, that's why i
was saying the rankings are using diff. criteria.
http://www.schooldigger.com/go/MD/schoolrank.aspx?level=3
even with diff. criteria, i'd say the real diff. is trival between these two.
centennial is 17 years older than river hill, and the community itself is
getting old too, not much new development is going on, and the location is
not as good as river hill, imho.
2010 |
|
f*****g
发帖数: 15860 | 43 hehe, if you never play a hand like this in 6-max, you'll be the sucker for
sure. if you play more hands like this, you'll be a sucker for sure too.
that's the reality in such games.
the point is, you have to adjust your aggression level in diff. games (9/10
regular, 6-max, 3 handed, heads up), againist diff. opponents, in diff.
modes (tilt, new to table, mix-up...). you play a sinlge style, you're
doomed to fail.
take this hand for example, both of them actually put the other one without
a stro |
|
l*t
发帖数: 10829 | 44 你们太重蓝轻绿了，为什么不想想俺家九姑娘是个胖妞呢?
俺开头就说了，九姑娘是瓜熟蒂落, 那意味着她至少跟她八哥一般重嘛.
九姑娘体重: 28磅
竟栽得泡屎[baozi]名单:
4楼: xianyunn, 28#
15楼: grapetomato, 29# （diff=1#)
17楼: creamer, 28#
18楼: liozodell 26.5# (diff=1.5#)
24楼: QUBO2011, 29.5# (diff=1.5#)
图1，图2: 称重真相
图3: 九姑娘跟哥哥姐姐的合影 |
|
a****a
发帖数: 26187 | 45 ☆─────────────────────────────────────☆
lct (红砖头) 于 (Thu Aug 4 21:12:52 2011, 美东) 提到:
今天下班推门进家，LD笑着跟俺说: 恭喜老爷，您中奖了，您的九千金瓜熟蒂落已入人间，快去看看她! ......
额地咯卖糕，昨天才得了八公子，今天又抱了九千金，介到底系好事还系坏事? 介么多
娃，俺咋养得起啊。
还别说，俺介九姑娘，天生的美绿，身材那个匀称，木得挑 - 见图1
## 竞猜完了，结果见44楼 ##
有奖竞猜庆祝九姑娘降生!
大家猜猜俺九姑娘体重是多少 (请精确到0.5磅)
规则有点变:
最先的头五个ID每人各得1包子 [误差须在2磅之内]
每人限猜两次,取最接近的
为避免不少童鞋像昨天那样的瞎猜，俺把俺家其他六个娃秀一下(见图2)
三愣子,四千金,五混子,六小姐,七狗剩, 八公子
三愣子最重，35磅
七狗剩最轻，15磅
九姑娘奏介于他们之间呀!
☆─────────────────────────────────────☆
liozodell (山水白鹤) 于 (Thu Aug 4... 阅读全帖 |
|
k***r
发帖数: 4260 | 46 Oh, sorry for being unclear.
We use ClearCase plug-in. The plug-in provides a diff window
to show differences between two revisions. The diff window
can't open. And I see some error messages from the console.
The same diff window opens fine on Windows.
Yeah, I'm going to try Ubuntu 64bit next. |
|
p*****3
发帖数: 488 | 47 原来java里有balanced的BST, 比C++的好用些，就是接口太多了，记不住。
写了一个求2d overlap矩形面积的题，
核心代码没多少，code都是定义各种结构各种override各种comparator去了:
public class EPI_14_20_3 {
static class Rectangle {
public int xBeg;
public int xEnd;
public int yBeg;
public int yEnd;
public Rectangle(int xb, int xe, int yb, int ye) {
xBeg = xb;
xEnd = xe;
yBeg = yb;
yEnd = ye;
}
}
static class EndPoint {
public int index... 阅读全帖 |
|
f*****t
发帖数: 895 | 48 自带的diff是不行。都是git可使用其他diff啊。喜欢什么diff拿来用就是了 |
|
l**********n
发帖数: 8443 | 49 Why do we need pure functions, immutable data and virtual DOM? These are
optimizations and to some extent simplifications, not core to idea.
var root = document.getElementById('ui');
var prevState = state, prevTree = [];
function render(state) {
// Virtual DOM is really just a tree of JavaScript objects or arrays
return [
['span', {id: 'count'}, state.items.length],
['ul', {}, state.items.map(function (item) {
return ['li', {}, item]
})]
]
}
funct... 阅读全帖 |
|
