bash leetcode
拓展:grep
193. ref: https://blog.csdn.net/yanglingwell/article/details/82343407
Given a text file file.txt
that contains list of phone numbers (one per line), write a one liner bash script to print all valid phone numbers.
You may assume that a valid phone number must appear in one of the following two formats: (xxx) xxx-xxxx or xxx-xxx-xxxx. (x means a digit)
You may also assume each line in the text file must not contain leading or trailing white spaces.
Example:
Assume that file.txt
has the following content:
987-123-4567
123 456 7890
(123) 456-7890
Your script should output the following valid phone numbers:
987-123-4567
(123) 456-7890
grep -P '^(\d{3}-|\(\d{3}\) )\d{3}-\d{4}$' file.txt
# . ^ 表示需要在行的开始处进行匹配. 例如, ^abc 可以匹配 abc123 但不匹配 123abc.
# . $ 表示需要在行的末端进行匹配. 例如, abc$ 可以匹配 123abc 但不能匹配 abc123.
# . \d可以匹配一个数字.'00\d'可以匹配'',但无法匹配'00A'.
# . {n}表示n个字符,用{n,m}表示n-m个字符. \d{}表示匹配3个数字,例如''.
# . -P(grep): Interpret the pattern as a Perl-compatible regular expression (PCRE).
195. ref: https://blog.csdn.net/sole_cc/article/details/44977821
Given a text file file.txt
, print just the 10th line of the file.
Example:
Assume that file.txt
has the following content:
Line 1
Line 2
Line 3
Line 4
Line 5
Line 6
Line 7
Line 8
Line 9
Line 10
Your script should output the tenth line, which is:
Line 10 Note:
1. If the file contains less than 10 lines, what should you output?
2. There's at least three different solutions. Try to explore all possibilities. 方法一:
awk NR== file.txt
//awk的默认动作就是打印$0,所以NR==10后面可以不用加{print $0}
方法二:
sed -n '10p' file.txt
//如果不够10行,则什么也不打印
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