连接suspect_id（而不是crime_id）上的两个表，得到句子开头和结尾之间的差，然后按句子长度的降序对表进行排序，并获取第一行（带有连接）以找到最大值：

SELECT s.name,
       FLOOR(MONTHS_BETWEEN(c.end_date, c.start_date) / 12) AS years,
       FLOOR(MOD(MONTHS_BETWEEN(c.end_date, c.start_date), 12)) AS months,
       31 * MOD(MONTHS_BETWEEN(c.end_date, c.start_date), 1) AS days
FROM   crime c
       INNER JOIN suspects s
       ON c.suspect_id = s.suspect_id
ORDER BY
       MONTHS_BETWEEN(c.end_date, c.start_date) DESC
FETCH FIRST ROW WITH TIES;

如果您使用的是Oracle 11或更早版本，则可以使用RANK分析函数：

SELECT *
FROM   (
  SELECT s.name,
         FLOOR(MONTHS_BETWEEN(c.end_date, c.start_date) / 12) AS years,
         FLOOR(MOD(MONTHS_BETWEEN(c.end_date, c.start_date), 12)) AS months,
         31 * MOD(MONTHS_BETWEEN(c.end_date, c.start_date), 1) AS days,
         RANK() OVER (ORDER BY MONTHS_BETWEEN(c.end_date, c.start_date) DESC) AS rnk
  FROM   crime c
         INNER JOIN suspects s
         ON c.suspect_id = s.suspect_id
)
WHERE  rnk = 1;