Adjacent Repeat of Numbers in an Array

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Olivia Rose
Olivia Rose 2022 年 4 月 12 日
コメント済み: Jacob 2023 年 4 月 19 日
This is the current function
function adjacentRepeat = HasAdjacentRepeat(inArray)
for inArray=diff(inArray)
adjacentRepeat=1;
end
It works to satisfy the following requirements:
  • Check for the use of a loop.
  • Assessment result: correctCheck adjacentRepeat = 1 for inArray = [2 1 3 3 4 2 5].
  • Assessment result: correctCheck adjacentRepeat = 1 for inArray = [4 5 7 6 3 8 9 7 2 8 9 7 3 3 3 7 1].
  • Assessment result: correctCheck for correct adjacentRepeat for an array of randomized size and values.
But not for this requirement:
  • Assessment result: incorrectCheck adjacentRepeat = 0 for inArray = [1 2 1 2 1].

採用された回答

Image Analyst
Image Analyst 2022 年 4 月 12 日
You never set adjacentRepeat to 0, EVER. So of course it doesn't work.
Did you try just going through your difference array checking if any are 0? If they are, then the adjacent values are the same and you can set the variable and exit.
function adjacentRepeat = HasAdjacentRepeat(inArray)
adjacentRepeat = 0;
d = diff(inArray);
for k = 1 : length(d)
if d(k) == 0
adjacentRepeat = 1;
return;
end
end
end
  2 件のコメント
Olivia Rose
Olivia Rose 2022 年 4 月 12 日
I attempted setting it to 0 and it wasn't working. Thank you for your help.
Jacob
Jacob 2023 年 4 月 19 日
I realized the problem I was having is I didnt understand what was or how to use diff. Thank you, this worked perfectly!

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その他の回答 (1 件)

Image Analyst
Image Analyst 2022 年 4 月 12 日
Why do you have a for loop? Why not simply do this:
adjacentRepeat = HasAdjacentRepeat([2 1 3 3 4 2 5])
adjacentRepeat = HasAdjacentRepeat([4 5 7 6 3 8 9 7 2 8 9 7 3 3 3 7 1])
adjacentRepeat = HasAdjacentRepeat([1 2 1 2 1])
function adjacentRepeat = HasAdjacentRepeat(inArray)
adjacentRepeat = any(diff(inArray) == 0);
end
  1 件のコメント
Olivia Rose
Olivia Rose 2022 年 4 月 12 日
It has to have a for loop

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