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8.24 Google SE OA 挂经,一共两题1小时完成。

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wxgymsfd 发表于 2016-9-22 12:52:09 | 显示全部楼层 |阅读模式

2016(7-9月) 码农类 硕士 全职@Google - Other - 其他 |Failfresh grad应届毕业生

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.鐣欏璁哄潧-涓浜-涓夊垎鍦

第一题:
You are given an integer X. You must choose two adjacent digits and replace them with the larger of these two digits.
For example, from the integer X = 233614, you can obtain:
33614 (by replacing 23 with 3);. 涓浜-涓夊垎-鍦帮紝鐙鍙戝竷
23614 (by replacing 33 with 3 or 36 with 6);
23364 (by replacing 61 with 6 or 14 with 4);
You want to find the smallest number that can be obtained from X by replacing two adjacent digits with the larger of the two. In the above example, the smallest such number is 23364.
Write a function:
class Solution {public int solution (int X);}
that, given a positive integer X, returns the smallest number that can be obtained from X by replacing two adjacent digits with the larger of the two.
For example, given X = 233614, the function should return 23364, as explained above.
Assume that:
X is an integer within the range [10..1,000,000,000].In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment.. 1point3acres.com/bbs
第二题You are given a listing of directories and files in a file system. Each directory and file has a name, which is a non-empty. more info on 1point3acres.com
string consisting of alphanumerical characters. Additionally, the name of each file contains a single dot character; the part.鏈枃鍘熷垱鑷1point3acres璁哄潧
of the name starting with the dot is called the extension. Directory names do not contain any dots. All the names are case-
sensitive.
Each entry is listed on a separate line. Every directory is followed by the listing of its contents indented by one space
character. The contents of the root directory are not indented.
Here is a sample listing:. From 1point 3acres bbs
dir1
dir11
dir12
  picture.jpeg
  dir 121.鐣欏璁哄潧-涓浜-涓夊垎鍦
   file1.txt
dir2
file2.gifWe have three files (picture.jpeg, file1.txt and file2.gif) and six directories (dir1, dir11, dir12, dir121, dir2 and the
root directory). Directory dir12 contains two files (picture.jpeg and file1.txt) and an empty directory (dir121). The root. From 1point 3acres bbs
directory contains two directories (dir1 and dir2).
The absolute path of a directory is a string containing the names of directories which have to be traversed (from the root
directory) to reach the directory, separated by slash characters. For example, the absolute path to the directory dir121 is
dir1/dir12/dir121". Note that there is no "drive letter", such as "C: ", and each absolute path starts with a slash. The. visit 1point3acres.com for more.
absolute path of a root directory is The image files are the files with extensions .jpeg, .png or .gif (and only these extensions). We are only interested in directories that directly contain image files (that is, without looking in their subdirectories). We want to find the total length
of the absolute paths to all the directories that directly contain at least one image file. For example, in the file system
described above there are two directories that directly contain image files: "/dir1/dir12" and "Mia". Directory "/dir1"
doesn't directly contain any image files. The total length of the absolute paths "/dir1/dir12" and "/dir2" is 11 + 5 = 16..鐣欏璁哄潧-涓浜-涓夊垎鍦
Write a function:
int solution(char *S);
that, given a string S consisting of N characters which contains the listing of a file system, returns the total of lengths (in
characters) modulo 1,000,000,007 of the absolute paths to all the directories directly containing at least one image file. For
example, given the sample listing shown above, the function should return 16, as explained above. If there are no image
files in the listing, the function should return 0.
Assume that:
N is an integer within the range [1..3,000,000];
string S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9), spaces, dots (.) and end--google 1point3acres
of-line characters;. from: 1point3acres.com/bbs
string S is a correct listing of a file system contents.
Complexity:
expected worst-case time complexity is 0(N);.1point3acres缃
expected worst-case space complexity is 0(N) (not counting the storage required for input arguments).

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