How to manipulate arrays?
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Hello! I am trying to create a script in which I sort an array in descending order without using any sort function. So far, I came up with this:
A = [2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19];
n = 1;
while n <= length(A)
if A(n) < A(n+1)
tmp = A(n);
A(n) = A(n+1);
A(n+1) = tmp;
end
n = n +1;
end
I thought it would be this simple, but it apparently isn't. One of the things I do not know how to do is place a value of one array into a new array, which I feel is vital in a script like this, but I did not know how to do it, so I did not implement it. Any help will be appreciated. Thank you in advance!
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その他の回答 (2 件)
Walter Roberson
2015 年 10 月 31 日
B(m) = A(n);
for some m and n.
However, that is not the reason your code is not working. You have implemented one pass of a "bubble sort". If your input were [3 2 1] then when n=1 it would change to [2 3 1] and then when n=2 it would change to [2 1 3] (and in the next pass it would crash because you would be trying to compare A(3) to the non-existent A(4), but that is trivially fixed.) Notice that the "2" and the "1" never get compared to each other. You will need to re-examine your strategy to fix that.
1 件のコメント
Shawn Simon
2015 年 10 月 31 日
編集済み: Shawn Simon
2015 年 10 月 31 日
timo
2015 年 10 月 31 日
Dont reinvent the wheel
>> x =py.list([5,4,1,2,3,7])
x =
Python list with no properties.
[5.0, 4.0, 1.0, 2.0, 3.0, 7.0]
>> x.sort
>> x
x =
Python list with no properties.
[1.0, 2.0, 3.0, 4.0, 5.0, 7.0]
3 件のコメント
Star Strider
2015 年 10 月 31 日
‘I am trying to create a script in which I sort an array in descending order without using any sort function.’
timo
2015 年 11 月 1 日
but why ? Too much free time ?
Star Strider
2015 年 11 月 1 日
It could be just to see how a sort script works, and to learn programming processes and logic in the process. Some people have the sort of insatiable curiosity about such things that will carry them far. Shawn is likely one such person.
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