面試題思路 - 台部落

cat words.txt | tr ' ' '\n' | grep -E -v '^$' | sort | uniq -c | sort -n -r | awk '{print $2,$1}'
cat words.txt | sed -e 's/^[ ]*//' -e 's/[ ]*$//' -e 's/[ ][ ]*/\n/g' | sort | uniq -c | sort -n -r | awk '{print $2,$1}'

193 有效電話號碼

from: https://leetcode-cn.com/problems/valid-phone-numbers/

思路：grep和正則表達式

cat file.txt | grep -E '(^$[0-9]{3}$ [0-9]{3}-[0-9]{4}$)|(^[0-9]{3}-[0-9]{3}-[0-9]{4}$)'

194 轉置文件

from: https://leetcode-cn.com/problems/transpose-file/

思路：awk運用或單行for循環

num=`head -1 file.txt | awk '{print NF}'`;for i in `seq $num`;do cut -d ' ' -f $i file.txt | tr '\n' ' ' | sed 's/ $/\n/';done
awk '{for(i=1;i<=NF;i++)if(NR==1){rst[i]=$i}else{rst[i]=rst[i]" "$i}} END {for(i=1;i<=NF;i++)print rst[i]}' file.txt
awk '{for(i=1;i<=NF;i++)rst[i]=rst[i]" "$i} END {for(i=1;i<=NF;i++)print substr(rst[i],2)}' file.txt

195 第十行

from: https://leetcode-cn.com/problems/tenth-line/

思路：文件遍歷行

sed -n '10p' file.txt
head -10 file.txt | tail -n +10
i=1;while read line;do if (( $i == 10 )); then echo $line; fi; i=$(($i+1));done < file.txt

數據庫

175 組合兩個表

from: https://leetcode-cn.com/problems/combine-two-tables

思路：left join用法

select a.FirstName, a.LastName, b.City, b.State from Person as a left join Address as b on a.PersonId = b.PersonId;

176 第二高的薪水

from: https://leetcode-cn.com/problems/second-highest-salary/

思路：limit用法，或者邏輯轉換：第二高的薪水，即比最高薪水小的最大薪水，使用臨時表求解，或者邏輯轉換：第二高的薪水，即比它高的薪水去重後個數等於1，使用子查詢求解。另外，答案外加select是爲了當沒有結果輸出時仍然輸出null結果。

select (select e.Salary from Employee as e group by Salary order by Salary desc limit 1,1) as SecondHighestSalary;
select max(a.Salary) SecondHighestSalary from Employee as a, (select max(Salary) as maxSalary from Employee) as b where a.Salary < b.maxSalary;
select (select distinct a.Salary from Employee as a where (select count(distinct b.Salary) from Employee as b where b.Salary > a.Salary) = 1) as SecondHighestSalary;

177 第N高的薪水

from: https://leetcode-cn.com/problems/nth-highest-salary/solution/jie-da-di-ngao-xin-shui-si-lu-by-joalpeng-yi-po/

思路：存儲過程中IF，變量和計算表達式等和limit用法

178 分數排名

from: https://leetcode-cn.com/problems/rank-scores/

思路：在查詢結果裏使用查詢語句，使用查詢語句構造臨時表

select a.Score as Score, (select count(b.distinctScore) + 1 from (select distinct(Score) as distinctScore from Scores) as b where b.distinctScore > a.Score) as Rank from Scores as a order by a.Score desc;

180 連續出現的數字

from: https://leetcode-cn.com/problems/consecutive-numbers/

思路：轉換邏輯：連續出現的數字即Id遞增，可使用多表條件查詢

select distinct a.Num as ConsecutiveNums from Logs as a, Logs as b, Logs as c where a.Num = b.Num and b.Num = c.Num and a.Id + 1 = b.Id and b.Id + 1 = c.Id;

181 超過經理收入的員工

from: https://leetcode-cn.com/problems/employees-earning-more-than-their-managers/

思路：基礎的多表條件查詢

select a.Name as Employee from Employee as a, Employee as b where a.ManagerId = b.Id and a.Salary > b.Salary;

182 查找重複的電子郵箱

from: https://leetcode-cn.com/problems/duplicate-emails/

思路：基礎的多表條件查詢

select distinct a.Email as Email from Person as a, Person as b where a.Id != b.Id and a.Email = b.Email;

183 從不訂購的客戶

from: https://leetcode-cn.com/problems/customers-who-never-order/

思路：基礎的子查詢

select a.Name as Customers from Customers as a where a.Id not in (select CustomerId from Orders);

184 部門工資最高的員工

from: https://leetcode-cn.com/problems/department-highest-salary/

思路：多表查詢和子查詢

select b.Name as Department, a.Name as Employee, a.Salary as Salary from Employee as a, Department as b where a.DepartmentId = b.Id and a.Salary = (select max(c.Salary) from Employee as c where c.DepartmentId = a.DepartmentId);

185 部門工資前三高的所有員工

from: https://leetcode-cn.com/problems/department-top-three-salaries/

思路：轉換邏輯：“工資前三高”即“比它高的工資去重後小於等於2”

select b.Name as Department, a.Name as Employee, a.Salary as Salary from Employee as a, Department as b where a.DepartmentId = b.Id and (select count(distinct c.Salary) from Employee as c where c.DepartmentId = a.DepartmentId and c.Salary > a.Salary) < 3;

196 刪除重複的電子郵箱

from: https://leetcode-cn.com/problems/delete-duplicate-emails/

思路：被刪除的表不能用於子查詢，故需要建立臨時表

delete from Person where Id in (select b.Id from (select * from Person) as b, (select * from Person) as c where b.Id > c.Id and b.Email = c.Email);

197 上升的溫度

from: https://leetcode-cn.com/problems/rising-temperature/

思路：DATEDIFF函數

select b.Id as Id from Weather as a, Weather as b where DATEDIFF(b.RecordDate, a.RecordDate) = 1 and b.Temperature > a.Temperature;

262 行程和用戶

from: https://leetcode-cn.com/problems/trips-and-users/

思路：查詢結果使用查詢語句，round函數

select a.Request_at as Day, round( (select count(*) from Trips as b where b.Request_at = a.Request_at and b.Status != 'completed' and b.Client_Id in (select Users_Id from Users where Banned = 'No') and b.Driver_Id in (select Users_Id from Users where Banned = 'No')) / count(*), 2) as 'Cancellation Rate' from Trips as a where a.Request_at <= '2013-10-03' and a.Request_at >= '2013-10-01' and a.Client_Id in (select Users_Id from Users where Banned = 'No') and a.Driver_Id in (select Users_Id from Users where Banned = 'No') group by a.Request_at;

595 大的國家

from: https://leetcode-cn.com/problems/big-countries/

思路：HelloWord類型

select name, population, area from World as a where a.area > 3000000 or a.population > 25000000;

596 超過5名學生的課

from: https://leetcode-cn.com/problems/classes-more-than-5-students/

思路：子查詢和count distinct

select distinct a.class as class from courses as a where (select count(distinct b.student) from courses as b where b.class = a.class) >= 5;

601 體育館的人流量

from: https://leetcode-cn.com/problems/human-traffic-of-stadium/

思路：基本的多表條件查詢，and/or的使用

select distinct a.id as id, a.visit_date as visit_date, a.people as people from stadium as a, stadium as b, stadium as c where a.people >= 100 and b.people >= 100 and c.people >= 100 and ((a.id + 1 =b.id and a.id + 2 = c.id) or (a.id -1 = b.id and a.id + 1 = c.id) or (a.id -2 = b.id and a.id -1 = c.id)) order by a.id;

620 有趣的電影

from: https://leetcode-cn.com/problems/not-boring-movies/

思路：基礎的條件查詢和order by

select * from cinema where description != 'boring' and id % 2 = 1 order by rating desc;

626 換座位

from: https://leetcode-cn.com/problems/exchange-seats/

思路：使用多表條件查詢，或者left join配合IF函數

select a.id as id, b.student as student from seat as a, seat as b, (select max(id) as maxId from seat) as c where (a.id % 2 = 1 and a.id + 1 = b.id) or (a.id % 2 = 0 and a.id - 1 = b.id) or (a.id %2 = 1 and a.id = c.maxId and b.id = a.id) order by a.id;
select a.id as id, IF(a.id % 2 = 1, IF(a.id = (select max(id) from seat), a.student, b.student), c.student) as student from seat as a left join seat as b on a.id + 1 = b.id left join seat as c on a.id -1 = c.id order by a.id;

627 交換工資

from: https://leetcode-cn.com/problems/swap-salary/

思路：IF函數

update salary set sex = IF(sex = 'm', 'f', 'm');

1179 重新格式化部門表

from: https://leetcode-cn.com/problems/reformat-department-table/

思路：group by裏對SUM和IF函數的運用

SELECT id, sum(IF(month='Jan',revenue,NULL)) Jan_Revenue, sum(IF(month='Feb',revenue,NULL)) Feb_Revenue, sum(IF(month='Mar',revenue,NULL)) Mar_Revenue, sum(IF(month='Apr',revenue,NULL)) Apr_Revenue, sum(IF(month='May',revenue,NULL)) May_Revenue, sum(IF(month='Jun',revenue,NULL)) Jun_Revenue, sum(IF(month='Jul',revenue,NULL)) Jul_Revenue, sum(IF(month='Aug',revenue,NULL)) Aug_Revenue, sum(IF(month='Sep',revenue,NULL)) Sep_Revenue, sum(IF(month='Oct',revenue,NULL)) Oct_Revenue, sum(IF(month='Nov',revenue,NULL)) Nov_Revenue, sum(IF(month='Dec',revenue,NULL)) Dec_Revenue FROM Department group by id;

多線程

1114 按序打印

from: https://leetcode-cn.com/problems/print-in-order/solution/javayou-jie-by-no-one-9/
https://leetcode-cn.com/problems/print-in-order/solution/an-xu-da-yin-by-leetcode/
https://leetcode-cn.com/problems/print-in-order/solution/javabing-fa-gong-ju-lei-da-lian-bing-by-kevinbauer/

思路：1）synchronized對對象加鎖，加鎖區域操作作爲標記的普通變量，退出加鎖區域前調用lock.notifyAll()喚醒其餘線程；2）使用AtomicInteger；3）無鎖方案，volatile和int賦值操作（int讀寫是原子操作，i++不是）4）其他：CountDownLatch的使用；Semaphore的使用；

1115 交替打印FooBar

from: https://leetcode-cn.com/problems/print-foobar-alternately/solution/wu-suo-qie-zui-jian-dan-zui-rong-yi-li-jie-de-shi-/

思路：類似上題，無鎖方案中，死循環裏要使用Thread.yield()以避免循環一直佔用CPU導致超時

1116 打印零與奇偶數

from: https://leetcode-cn.com/problems/print-zero-even-odd/solution/wu-xu-jia-suo-ding-yi-intlei-xing-flagshi-yong-012/

思路：類似上題，無鎖方案裏，volatile int的值使用0，1，2表示3種狀態。

1117 H2O 生成

from: https://leetcode-cn.com/problems/building-h2o/solution/xin-hao-liang-by-hansin1997-3/

思路：Semaphore用法

1195 交替打印字符串

from: https://leetcode-cn.com/problems/fizz-buzz-multithreaded/solution/liang-chong-javajie-jue-fang-an-shi-yong-semaphore/
https://leetcode-cn.com/problems/fizz-buzz-multithreaded/solution/1ge-reentrantlock-1ge-condition-1ge-volatilebian-l/

思路：Semaphore用法或者無鎖方案

1226 哲學家進餐

from: https://leetcode-cn.com/problems/the-dining-philosophers/solution/1ge-semaphore-1ge-reentrantlockshu-zu-by-gfu/

思路：避免死鎖的幾種策略見下，使用Semaphore，ReentrantLock，synchronized等實現

1）哲學家0先爭用左手叉，再爭用右手叉；其他哲學家先爭用右手叉，再爭用左手叉。（所以，哲學家0，1會爭用它們之間的叉子，肯定有一個爭不到，避免了5人同時進餐的死鎖情況）
2）設置信號量最多運行4個哲學家同時進食（包括選擇叉子-吃-放下叉子），並且對每個叉子設置信號量防止被2個哲學家同時使用。
3）設置信號量最多運行4個哲學家同時選擇叉子，並且對每個叉子設置信號量防止被2個哲學家同時使用。
4）選擇叉子時設置臨界區，拿到2個叉子後再退出臨界區（同一時刻只有1個哲學家可以選擇叉子，防止全部人搶叉子造成死鎖），並且對每個叉子設置信號量防止被2個哲學家同時使用。
5）設置臨界區每次只讓1個哲學家進食（包括選擇叉子-吃-放下叉子），此時無需對叉子進行限制。