Lintcode logo queries the two nearest saplings

The requirement of the question is the distance between the nearest two trees in the query table . And rename the result （shortest_distance）

Answer 1 ：（ Adopt aggregate function and self connection ）

SELECT MIN(ABS(a.distance - b.distance)) AS shortest_distance
FROM sapling_distances AS a
INNER JOIN sapling_distances b
ON a.id != b.id;

This is a very simple solution , Fields are not complicated .
Attention is paid to small details . Distance , The preconditions for finding the absolute value and the added minimum value are adopted . Then there is self join query . The condition of self connection query is the combination of both id inequality . If they are the same, then the query distance is meaningless . If id The same is comparing itself . What is the analogy between oneself and oneself ？
！= You can also use <> Instead of . It means the same thing

Explanation 2 ：（ Adopt nested query thinking ）

SELECT MIN(distance_diff) AS shortest_distance FROM (
	   SELECT  abs(b.distance - a.distance) AS distance_diff FROM sapling_distances a, sapling_distances b
	   WHERE a.id <> b.id
) cc
HAVING shortest_distance is not null;

It adopts the thinking of nested query , It's also very easy to understand . The outer layer minimizes the required result , The inner layer calculates the absolute value of the query distance . The processing of the table in memory is to name the table twice , Make it a different watch , Then attach conditions . The last thing to note is cc This is the name of the sub query table , If not , Will report a mistake , This is the grammatical requirement .

Answer 3 ：（ Another mode of thinking , It's nothing special ）

select min(a.distance - b.distance) shortest_distance
from sapling_distances a
    join sapling_distances b on a.distance > b.distance
having shortest_distance is not null;

One limitation a.distance>b.distance This replaces abs（） Aggregate functions , In this way, you can also get normal results .

Solution 4 ：（ Superfluous solution , Simple problems complicate ）

select min(s2.distance-s1.distance) shortest_distance from
(select distance,@rownum1:=@rownum1+1 r1 
from sapling_distances,(select @rownum1:=0) ra
order by distance) s1,
(select distance,@rownum2:=@rownum2+1 r2 
from sapling_distances,(select @rownum2:=0) rb
order by distance) s2
where s1.r1 = s2.r2-1 having shortest_distance is not null;

Don't explain , Because I won't . Solution from netizens . I really hate this solution . Simple problems complicate , Don't do it . ok ！ I haven't used this method yet , When you learn it, you can add .