某公司数据库面试题

Question 1:
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写一个函数 把传入的数组倒置 可以用任何编程语言
不能用现有函数，除了coun或者size之类的基本操作

Question 2:
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In database XYZ, there's a table 'run_key' with the following rows:

idbatch last_processed
------------------------------
1  5           999
26           999
3  7             0
4  5             0
5 6           999
6  7             0
7  5             0
8  6             0
97           999
107             0


Write a SQL statement to return the number of rows with last_processed=0 for each batch.
The expected output is:

batchcount
-------------
5            2
6            1
7            3

Question 3:
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In table 'info', there's a column named 'graduation_date' of type date. Some rows have 
incorrect graduation_date such as '1907-11-09' and '1908-03-17'; they should be 
'2007-11-09' and '2008-03-17'. 
Write a SQL statement to update all the rows with graduation_date < '1909-01-01' by 
setting graduation_date to '20yy-mm-dd'.
For example, '1907-11-09' updated to '2007-11-09', and '1908-03-17' updated to '2008-03-17'.

Question 4:
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in case you are not familiar with regular expression matching, here's a quick example:

if we have a string "<year>2009</year>" and we want to extract '2009', we can use
/<year>(.*)</year>/.

based on the example, write ONE regular expression that will match "BALTIMORE COUNTY" in 
"<p><strong>BALTIMORE COUNTY</strong></p>" AND "NEW YORK"
in "<p><strong>NEW YORK</strong></p>"

Question 5:
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in case you are not familiar with regular expression substitution, here's a quick tutorial:

the syntax of string substitution is:
VARIABLE =~ s/SEARCH_PATTERN/NEW_STRING/

For example,
$a = 'abc,123';
$a =~ s/(w+),(w+)/2,1/; # $a is now '123,abc' because 1='abc' and 2='123'

Here's the question:
write ONE substitution statement(ie. s/SEARCH_PATTERN/NEW_STRING/) so that
"<date>1999-02-25</date>" will be updated to "<date>02-25-1999</date>" AND

"<date>2005-11-03</date>" will be updated to "<date>11-03-2005</date>"

Question 6:
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learn the concept of 'hash table' first. then solve this:

array 1 has some integers (for example: 1, 3, 5, 7) and array 2 has some integers (for example: 8, 5, 6, 1).

write a function to find the integers that exist in both arrays.

Question 7:
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mysql> select * from a;
+--------+
| letter |
+--------+
| x      |
| y      |
| z      |
+--------+
mysql> select * from b;
+--------+
| letter |
+--------+
| a      |
| b      |
| y      |
+--------+

write a query to return letters that exist in both table a and table b;
write a query to return letters that exist in table a but not in table b.

Question 8:
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写一个函数 传入一个数组和N 要求把前N个元素移到最后 需要占用最少内存

比如传入[a b c d e], N=2, 要求返回数组[c d e a b]

答案：（稍后）

answer1:

下面是我用java语言写的一维数组和二维数组的转置程序。
一维数组的转置：

package arraytest;
public class ArrayInversion1 {
public static void main(String[] args) {
int[] a={2,3,4,5};
arrayInversion(a);
for(int i:a){
System.out.println(i);
}
}
//一位数组倒置
public static void arrayInversion(int[] a){
int temp;
for(int i =0;i<=(a.length)/2;i++){
temp = a[i];
a[i]=a[a.length-i-1];
a[a.length-i-1] =temp;
}
}
}

二维数组的转置：

package arraytest;
public class ArrayInversion2 { 
    public static void main(String[] args) {  
     //二维数组可以不规则
   
        int a[][]={{1,2,3},{4,5,6},{7,8}};
        
        show(a);
     
        int[][] rea =arrayInversion(a);
          
        show(rea);
          
    }  
    
    
    //倒置方法
     public static int[][] arrayInversion(int[][] a){
     //创建临时变量count，记录数组中长度最大的数组length
        int count =0;
          
        for(int i = 0;i
       
        if(count
        count = a[i].length;
        }
        }
        
       // System.out.println("max"+count);
        
        //创建数组用来装载倒置的数组数据
        int rea[][]=new int[count][a.length];
       
        //倒置
        for(int i=0 ; i
            for(int j=0; j
                rea[j][i]=a[i][j];   
            }     
        }  
   
    return rea;
    }
    
     
     //输出数组
    public static void show(int[][] a){
        for (int x[]:a){  
            for(int e:x){  
                System.out.print(e+" ");  
            }  
            System.out.println();  
        }  
        System.out.println();  
    }
  
}  

answer2：

select batch,count(batch) as count from run_key where last_processed=0 group by batch;