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

# 1. ^ 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.
# 2. $ 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.
# 3. d可以匹配一个数字.'00d'可以匹配'007'，但无法匹配'00A'.
# 4. {n}表示n个字符，用{n,m}表示n-m个字符. d{3}表示匹配3个数字，例如'010'.
# 5. -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==10 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