How would you print just the 10th line of a file?
For 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
即输出第十行，又是bash题。
sed -n 10p file.txt
没什么可说的。题目要求3种以上解法。不过对于bash题，暂时以学习了解为主。
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.
For 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
即从文件中找到合法的电话号码。又进入bash题区了。
grep -P '^(\d{3}-|\(\d{3}\) )\d{3}-\d{4}$' file.txt
没啥可说的，正则题而已。唯一需要注意的是，这里用的是\d ，所以需要-P支持。
Write a function that takes an unsigned integer and returns the number of ’1’ bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3.
求整数转化为二进制数中，1的个数
# Write a function that takes an unsigned integer and returns the number of ’1' bits
# it has (also known as the Hamming weight).
#
# For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011,
# so the function should return 3.
# author li.hzh
class Solution(object):
def hammingWeight(self, n):
"""
:type n: int
:rtype: int
"""
result = 0
while n:
n = n & n - 1
result += 1
return result
print(Solution().hammingWeight(11))
其实完全可以通过Reverst Bit中的方法，通过 & 1，右移把1统计出来。不过效率会略低。因为每一位都要检查。
这里通过一个技巧 n & (n-1)，想象一下，每次都会消掉一个1，就加快了效率。
Reverse bits of a given 32 bits unsigned integer.
For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).
Follow up: If this function is called many times, how would you optimize it?
即将整数的32个bit，逆序后，输出对应的整数。
# Reverse bits of a given 32 bits unsigned integer.
#
# For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).
#
# Follow up:
# If this function is called many times, how would you optimize it?
# author li.hzh
class Solution:
# @param n, an integer
# @return an integer
def reverseBits(self, n):
result = 0
digits = 0
while digits < 32:
temp = n & 1
result <<= 1
result |= temp
n >>= 1
digits += 1
return result
solution = Solution()
print(solution.reverseBits(43261596))
通过位运算取整移位。不得不说，算是自己的一个进步吧，能应用位运算了。
Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
Note: Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
[show hint]
Hint: Could you do it in-place with O(1) extra space?
即按照要求移位数组。最好不利用额外空间。
# Rotate an array of n elements to the right by k steps.
#
# For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
#
# Note:
# Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
#
# [show hint]
#
# Hint:
# Could you do it in-place with O(1) extra space?
# Related problem: Reverse Words in a String II
# author li.hzh
class Solution:
def rotate(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: void Do not return anything, modify nums in-place instead.
"""
length = len(nums)
k = k % length
if length <= 1 or k == 0:
return
end_index = length - 1
for i in range(length):
temp = nums[i]
current_end = end_index - i
if i < current_end:
nums[i] = nums[current_end]
nums[current_end] = temp
else:
break
for i in range(k):
temp = nums[i]
current_end = k - i - 1
if i < current_end:
nums[i] = nums[current_end]
nums[current_end] = temp
else:
break
for i in range(k, length):
temp = nums[i]
current_end = end_index - i + k
if i < current_end:
nums[i] = nums[current_end]
nums[current_end] = temp
else:
return
solution = Solution()
nums = [1, 2, 3]
solution.rotate(nums, 1)
print(nums)
题目说用3种以上不同的解法。这里只给出了一种，O(n)时间，O(1) Space的，比较好理解的解法。就是 [1,2,3,4,5] ，先反转[5,4,3,2,1]，再按照k分割前后反转就是答案。比如k=2,[4,5,1,2,3]。代码很容易写。
如果使用额外空间，那么O(n)的解法也很直接。这里不再赘述。
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.
Table: Customers.
+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+
Table: Orders.
+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+
Using the above tables as example, return the following:
+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+
即找到没有订单的客户
SELECT Name AS Customers FROM Customers c WHERE c.Id NOT IN (SELECT CustomerId FROM Orders);
SELECT Customers FROM (SELECT c.Name AS Customers, o.Id AS oid FROM Customers c LEFT JOIN Orders o ON c.Id = o.CustomerId) AS r WHERE r.oid IS NULL;
无
Write a SQL query to find all duplicate emails in a table named Person.
+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+
For example, your query should return the following for the above table:
+---------+
| Email |
+---------+
| a@b.com |
+---------+
找到重复的Email。
SELECT Email FROM Person GROUP BY Email HAVING COUNT(Email) > 1;
无
The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
+----+-------+--------+-----------+
| Id | Name | Salary | ManagerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | NULL |
| 4 | Max | 90000 | NULL |
+----+-------+--------+-----------+
Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
+----------+
| Employee |
+----------+
| Joe |
+----------+
找到比Manager薪水高的雇员
SELECT e1.Name AS Employee FROM Employee e1 INNER JOIN Employee e2 ON e1.ManagerId = e2.Id WHERE e1.Salary > e2.Salary;
通过Join解决问题
Write a SQL query to get the nth highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
For example, given the above Employee table, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.
+------------------------+
| getNthHighestSalary(2) |
+------------------------+
| 200 |
+------------------------+
上一个问题的推广，找到第N大的元素
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
SET N = N - 1;
RETURN (
# Write your MySQL query statement below.
SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1 OFFSET N
);
END
解题思路一致，不过是改成个自定义函数而已。
Write a SQL query to get the second highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
For example, given the above Employee table, the query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+
即找到第二大的元素。没有返回null。
# Write your MySQL query statement below
SELECT (SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1 OFFSET 1) AS SecondHighestSalary;
通过LIMIT OFFSET找到第二个元素。DISTINCT是为了排除两个最大值的情况。嵌套是为了解返回null的情况。另一种办法是使用IFNULL函数。