alh*_*iya 3 sql oracle plsql subset-sum
我正在寻找你的专家的指导,找出一种方法来比较一个数字与数字子集的总和,如
DECLARE
L_NUM_TO_COMPARE NUMBER := 0;
L_NUM_SUBSET NUMBER := 0;
BEGIN
FOR MAIN_REC IN (
SELECT 1 ID, 25150 ASSIGN_AMT FROM DUAL
UNION ALL
SELECT 2 ID, 19800 ASSIGN_AMT FROM DUAL
UNION ALL
SELECT 3 ID, 27511 ASSIGN_AMT FROM DUAL
) LOOP
L_NUM_TO_COMPARE := MAIN_REC.ASSIGN_AMT;
DBMS_OUTPUT.PUT_LINE( L_NUM_TO_COMPARE);
FOR C IN (
SELECT 1 ID, 7120 WORK_AMT FROM DUAL
UNION ALL SELECT 2 ID, 8150 WORK_AMT FROM DUAL
UNION ALL SELECT 3 ID, 8255 WORK_AMT FROM DUAL
UNION ALL SELECT 4 ID, 9051 WORK_AMT FROM DUAL
UNION ALL SELECT 5 ID, 1220 WORK_AMT FROM DUAL
UNION ALL SELECT 6 ID, 12515 WORK_AMT FROM DUAL
UNION ALL SELECT 7 ID, 13555 WORK_AMT FROM DUAL
UNION ALL SELECT 8 ID, 5221 WORK_AMT FROM DUAL
UNION ALL SELECT 9 ID, 812 WORK_AMT FROM DUAL
UNION ALL SELECT 10 ID, 6562 WORK_AMT FROM DUAL
ORDER BY 2 DESC
) LOOP
L_NUM_SUBSET := NVL(L_NUM_SUBSET,0) + C.WORK_AMT;
DBMS_OUTPUT.PUT_LINE( L_NUM_SUBSET);
/*
I NEED TO PUT SOME LOGIC HOW CAN I FIND NEAREST SUM OF SUBSET
*/
IF MAIN_REC.ASSIGN_AMT = L_NUM_SUBSET THEN
DBMS_OUTPUT.PUT_LINE( L_NUM_SUBSET);
END IF;
END LOOP;
END LOOP;
END;
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我在这个论坛上搜索过,发现了一个问题 Sum of Sub数集
这几乎与我的要求相同,我需要的是一些人可以指出我如何在PL/SQL中做到这一点(Oracle DB 11g R2)
Luk*_*der 12
您不需要PL/SQL来解决这个问题.这是一个非常有趣的问题,单独用SQL来解决,我写了一篇博文来更详细地解释我的答案.
你提出问题的方式,我假设你并没有真正解决子集和问题,而是一个更简单的问题,你想要将一个数字与一组非常有限的子集进行比较,即按顺序排列的子集.WORK_AMT上升,没有差距.
这可以单独使用Oracle SQL解决:
WITH
ASSIGN(ID, ASSIGN_AMT) AS (
SELECT 1, 25150 FROM DUAL
UNION ALL SELECT 2, 19800 FROM DUAL
UNION ALL SELECT 3, 27511 FROM DUAL
),
VALS (ID, WORK_AMT) AS (
SELECT 1 , 7120 FROM DUAL
UNION ALL SELECT 2 , 8150 FROM DUAL
UNION ALL SELECT 3 , 8255 FROM DUAL
UNION ALL SELECT 4 , 9051 FROM DUAL
UNION ALL SELECT 5 , 1220 FROM DUAL
UNION ALL SELECT 6 , 12515 FROM DUAL
UNION ALL SELECT 7 , 13555 FROM DUAL
UNION ALL SELECT 8 , 5221 FROM DUAL
UNION ALL SELECT 9 , 812 FROM DUAL
UNION ALL SELECT 10, 6562 FROM DUAL
),
SUMS (ID, WORK_AMT, SUBSET_SUM) AS (
SELECT VALS.*, SUM (WORK_AMT) OVER (ORDER BY ID)
FROM VALS
)
SELECT
ASSIGN.ID,
ASSIGN.ASSIGN_AMT,
MIN (SUBSET_SUM) KEEP (
DENSE_RANK FIRST
ORDER BY ABS (ASSIGN_AMT - SUBSET_SUM)
) AS CLOSEST_SUM
FROM
ASSIGN
CROSS JOIN
SUMS
GROUP BY
ASSIGN.ID, ASSIGN.ASSIGN_AMT
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以上产量:
ID ASSIGN_AMT CLOSEST_SUM
---------------------------
1 25150 29085
2 19800 20935
3 27511 29085
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请注意,此问题在时间和空间上具有指数级复杂性.它只能合理地解决WORK表中的少量值!
WITH
ASSIGN (ID, ASSIGN_AMT) AS (
SELECT 1, 25150 FROM DUAL
UNION ALL SELECT 2, 19800 FROM DUAL
UNION ALL SELECT 3, 27511 FROM DUAL
),
WORK (ID, WORK_AMT) AS (
SELECT 1 , 7120 FROM DUAL
UNION ALL SELECT 2 , 8150 FROM DUAL
UNION ALL SELECT 3 , 8255 FROM DUAL
UNION ALL SELECT 4 , 9051 FROM DUAL
UNION ALL SELECT 5 , 1220 FROM DUAL
UNION ALL SELECT 6 , 12515 FROM DUAL
UNION ALL SELECT 7 , 13555 FROM DUAL
UNION ALL SELECT 8 , 5221 FROM DUAL
UNION ALL SELECT 9 , 812 FROM DUAL
UNION ALL SELECT 10, 6562 FROM DUAL
),
SUMS (SUBSET_SUM, MAX_ID) AS (
SELECT WORK_AMT, ID FROM WORK
UNION ALL
SELECT WORK_AMT + SUBSET_SUM, GREATEST (MAX_ID, WORK.ID)
FROM SUMS JOIN WORK
ON SUMS.MAX_ID < WORK.ID
)
SELECT
ASSIGN.ID,
ASSIGN.ASSIGN_AMT,
MIN (SUBSET_SUM) KEEP (
DENSE_RANK FIRST
ORDER BY ABS (ASSIGN_AMT - SUBSET_SUM)
) AS CLOSEST_SUM
FROM SUMS
CROSS JOIN ASSIGN
GROUP BY ASSIGN.ID, ASSIGN.ASSIGN_AMT
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这现在产生:
ID ASSIGN_AMT CLOSEST_SUM
---------------------------
1 25150 25133
2 19800 19768
3 27511 27488
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