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write a function called palindrome that takes one input argument a char vector and recursively determine whether that argument is a palindrome you are not allowed to use loops not built in function like srtcmp etc. the function returns true or false

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JAYANTHI SANKARALINGAM
回答： Tarun Sangwan 2024-3-27
function ok=palindrome(txt)
iflength(txt)<=1
ok=true;
else
ok=(txt(1)==txt(end)&&palindrome(txt(2:end-1)));
end
end
hi,
It shows error in else statment in the above program, kindly give me a solution to solve the above problem.

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采纳的回答

Ameer Hamza
Ameer Hamza 2020-10-21
编辑：Ameer Hamza 2020-10-21
There should be a space between if keyword and the condition
if length(txt)<=1
%^ insert a space here
Apart from that, the logic is correct, but you just have a misplaced bracket. Following is correct
ok=(txt(1)==txt(end))&&palindrome(txt(2:end-1));

更多回答（3 个）

Sandeep Kumar Patel
编辑：DGM 2024-1-10
If we apply the edits recommended by @Ameer Hamza, we get this:
function ok=palindrome(txt)
if length(txt)<=1 % added space
ok = true;
else
ok = (txt(1)==txt(end)) && palindrome(txt(2:end-1));
% close parentheses --^
end
end
Black Woods
Black Woods 2022-12-12
function ans=palindrome(v)
if v(1)==v(end)
ans=true;
if length(v)==1 || length(v)==2
return
else
palindrome(v(2:length(v)-1));
end
else
ans=false;
end
end
Tarun Sangwan
Tarun Sangwan 2024-3-27
function p = palindrome(n)
if length(n)/2<1
p = 2
else
p = n(1) == n(end)
p = [p palindrome(n(2:end-1))]
end
p = ~ismember(0,p)

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