我目前正在学习COBOL,并且正在尝试在我的程序中实现冒泡排序算法。虽然我对这门语言还很陌生,但我写的东西在语义和语法上对我来说是有意义的,但是如果我按这个顺序输入5,4,3,2和1,我的帖子排序表就变成了1,5,4,3,2。有人能解释一下我哪里错了吗?
IDENTIFICATION DIVISION.
PROGRAM-ID. BubbleSort.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 TVAR PIC 9(4).
01 CNT PIC 9(1) VALUE 1.
01 CNT2 PIC 9(1) VALUE 1.
01 ARR.
05 ARRELEMENT PIC 9(4) OCCURS 5 TIMES.
01 TABLELENGTH PIC 9(1) VALUE 5.
PROCEDURE DIVISION.
DISPLAY "Enter 5 numbers: ".
PERFORM INPUT-PARA VARYING CNT FROM 1 BY 1 UNTIL CNT>5.
DISPLAY "Pre Bubble-Sort: ".
PERFORM PRINT-PARA VARYING CNT FROM 1 BY 1 UNTIL CNT>5.
PERFORM BBLSORT-PARA.
DISPLAY "Post Bubble-Sort: ".
PERFORM PRINT-PARA VARYING CNT FROM 1 BY 1 UNTIL CNT>5.
STOP RUN.
INPUT-PARA.
ACCEPT ARRELEMENT(CNT).
PRINT-PARA.
DISPLAY "Table element: "ARRELEMENT(CNT).
BBLSORT-PARA.
INITIALIZE CNT CNT2.
MOVE 1 TO CNT.
MOVE 2 TO CNT2.
PERFORM UNTIL CNT>6
PERFORM UNTIL CNT2>5
DISPLAY "IF "ARRELEMENT(CNT) " IS > "ARRELEMENT(CNT2)
IF (ARRELEMENT((CNT)) > ARRELEMENT((CNT2)))
THEN
DISPLAY ARRELEMENT(CNT) " IS > "ARRELEMENT(CNT2)
MOVE ARRELEMENT(CNT) TO TVAR
MOVE ARRELEMENT(CNT2) TO ARRELEMENT(CNT)
MOVE TVAR TO ARRELEMENT(CNT2)
END-IF
DISPLAY "EXIT IF LOOP"
ADD 1 TO CNT2 GIVING CNT2
END-PERFORM
ADD 1 TO CNT GIVING CNT
END-PERFORM.
END PROGRAM BubbleSort.
发布于 2020-02-09 11:04:56
免责声明:我不是COBOL程序员,但通过对循环的这些更改,我能够做到这一点。
PERFORM VARYING CNT
FROM 1 BY 1
UNTIL CNT > 4
PERFORM VARYING CNT2
FROM 1 BY 1
UNTIL CNT2 + CNT > 5
COMPUTE
CNT3 = CNT2 + 1
END-COMPUTE
IF (ARRELEMENT((CNT2)) > ARRELEMENT((CNT3)))
DISPLAY ARRELEMENT(CNT2) " IS > "ARRELEMENT(CNT3)
MOVE ARRELEMENT(CNT3) TO TVAR
MOVE ARRELEMENT(CNT2) TO ARRELEMENT(CNT3)
MOVE TVAR TO ARRELEMENT(CNT2)
END-IF
DISPLAY "EXIT IF LOOP"
END-PERFORM
END-PERFORM.
关键是添加一个CNT3,我用它来代表CNT2 + 1。在内部循环之前,每次都需要将CNT2重新发送到1。然后,该算法在每次不引用CNT的情况下,在循环内比较项CNT2与项CNT2+1。
而且,不需要每次都在内部循环中一直走到数组的末尾。我发现参考Geeks for Geeks on bubble sort很有帮助。
https://stackoverflow.com/questions/60132893
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