Recursive function to reverse a vector
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The question is to write a function to reverse a vector so v=[1 2 3] becomes v= [3 2 1]
It must be recursive
So my idea how to write this function is start off with d=1 the last digit of w, the output, is equal to the first digit of v the input. so thats what I wrote. As the function develops d eventually equals the length of the original input symoblising the function has reached an end and it will return w. I just don't know how to keep d growing everytime, as currently with every time I recall the function d will rewrite itself to be 1. I think it has something to do with the base case but I don't know. I'm not looking for the answer on a plate but more like a discussion of someone talking me through it
function [w]=reversal(v)
d=1
e=length(v)
w(end-d+1)=v(d)
if d==e
w=w
else
[w]=reversal(v)
end
end
採用された回答
その他の回答 (2 件)
David Hill
2020 年 10 月 21 日
編集済み: David Hill
2020 年 10 月 21 日
function w=reversal(v)
if length(v)==1
w= v;
else
w= [v(end),reversal(v(1:end-1))];
end
3 件のコメント
katherine keogh
2020 年 10 月 21 日
David Hill
2020 年 10 月 21 日
Just like John explained below. You stop the recursion when the length of the vector is equal to 1, otherwise you keep calling the function and growing the layers of recursion. For example: [1 2 3 4] becomes
w=[4 , reversal([1 2 3])]
w=[3, reversal([1 2])]
w=[2, reversal([1])]
w=1
You go 4 layers into the recursion. When the recursion ends, it all comes together.
w=[4 3 2 1]
katherine keogh
2020 年 10 月 21 日
SUTITHI
2022 年 11 月 22 日
0 投票
function w=reversal(v) if length(v)==1 w= v; else w= [v(end),reversal(v(1:end-1))]; end
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