Updated code to suit your requirements:

WITH RankedAccounts AS (
  SELECT
    ds.cust_id,
    dt.acct_no AS curr_loan,
    ds.acct_no AS prev_loan,
    ROW_NUMBER() OVER (PARTITION BY ds.cust_id ORDER BY ds.date_last_disb DESC) AS RowNum
  FROM acct_dtls dt
  INNER JOIN acct_dtls ds
    ON (dt.cust_id = ds.cust_id)
  WHERE dt.date_last_disb IS NOT NULL
    AND ds.date_last_disb < dt.date_last_disb
)

SELECT cust_id, curr_loan, prev_loan
FROM RankedAccounts
WHERE RowNum = 1;

对于每个客户，您需要一个包含最终和倒数第二个贷款的结果行。实现此目的的一种方法是使用LAG获取行的precedent，使用LEAD获取行的successor：

SELECT
  cust_id,
  acct_no as current_loan, 
  prev_acct_no as previous_loan
FROM
(
  SELECT
    cust_id,
    acct_no,
    LAG(acct_no) OVER (PARTITON BY cust_id ORDER BY date_last_disb) AS prev_acct_no,
    LEAD(acct_no) OVER (PARTITON BY cust_id ORDER BY date_last_disb) AS next_acct_no
  FROM acct_dtls
  WHERE date_last_disb IS NOT NULL
) with_prev_and_next
WHERE next_acct_no IS NULL
ORDER BY cust_id;