l***h
发帖数: 96 | 1 上上周五去MTV onsite的，这周一HR说HC已经过了，等这周EC省材料，下周给我消息。
希望不要在最后一步出啥问题吧。
电面题目：
一位国人大哥，先在这里谢过啦，时间刚好45分钟，吐槽下Google docs上写程序好蛋
疼，什么时候可以搞个可以支持Vim的编辑器吧。。。。
Assume some binary (each pixel is either black or white ) images, have same
width and height, and the length is power of 2. Present it by quadtree:
divide the image into quarters, each quarter can be all back, all white or
mixed, subdivide the mixed ones… recurse.
+-------+---+---+
| | F | F |
| T +---+---+
| | T | T |
+---+---+---+---+
| F ... 阅读全帖 |
|
f******h
发帖数: 45 | 2 也找工作了一段时间了，从版上学了很多，上周G家面完了，求个bless。
之前的一些都挂了，还在继续找其他的。等定下来之后一定发面经回报本版。
谢谢大家啦！！
1. http://www.mitbbs.com/article_t/JobHunting/32005597.html
1) Implement a simple calculator (+,-,*,/);
2) Implement "+1" for a large integer;
3) How to match Ads to users;
4) How to extract useful information from a forum webpage (list all
kinds of useful signal you can think of)
5) How to detect the duplicate HTML pages (large scale);
6) Find all the paths between two places on Google map;
7)... 阅读全帖 |
|
w**7
发帖数: 22 | 3 b) count all the black pixels of this image
int getBlackPixels( TreeNode* root) {
if(!root)
return 0;
if( !root->upperLeft) {
if( root->pixel)
return root->size * root->size;
}
int sum = 0;
sum += getBlackPixels( root->upperLeft);
sum += getBlackPixels( root->downLeft);
sum += getBlackPixels( root->upperRight);
sum += getBlackPixels( root->downRight);
return sum;
}
The code above does not look right. A node will have its pixel size ... 阅读全帖 |
|
b*******r
发帖数: 432 | 4 check www.metroairport.com, then on the Site Menu (upperleft corner)
choose "Maps". |
|
t*********i
发帖数: 217 | 5 since you are using toad, find SYS in the drop down list on upperleft side,
click it; find the package or procedure label on left panel, click it, then
find the procedure.packages you want to see.
Ask your DBA for privilege if you don't have the permission to view sys/
system objects.
But I doubt they will give it to you if you are not a DBA... |
|
n*******r
发帖数: 444 | 6 Divide & Conquer应该可以解决这个问题．把矩阵一分为二，递归地解决每一个
sub-matrix, 再考虑across sub-matrix border的方案，这应该是一个
O(N^2lgN)的算法．
另外一个方法是对每一个black cell分类，即：可以为lowerLeft, upperRight,
lowerRight and upperLeft的cell,然后对他们做配对处理，（可以考虑把它
转成图的问题，即求有向图最大的circle). |
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