(1)Back and Forth Rows
Given a number n, create an n-by-n matrix in which the integers from 1 to n^2 wind back and forth along the rows as shown in the examples below.
Examples:
Input n = 3
Output a = [ 1 2 3
6 5 4
7 8 9 ]
Input n = 4
Output a = [ 1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13 ]
答：
b = zeros(n);
b(:) = (1:n.^2);
c = flipud(b);
b(:,2:2:size(b,2)) = c(:,2:2:size(b,2));
b=b';
end
答：
x=reshape(1:n*n,n,n)';
for i =2:2:n
x(i,:)=flip(x(i,:));
end
b=x;
虽然第二种方法有循环，但是速度更快。
(2)Most nonzero elements in row
Given the matrix a, return the index r of the row with the most nonzero elements. Assume there will always be exactly one row that matches this criterion.
Example:
Input a = [ 1 2 0 0 0
0 0 5 0 0
2 7 0 0 0
0 6 9 3 3 ]
Output r is 4
答：
[x,r]=min(sum(a==0,2));
答
[x,r]=max(sum(a~=0,2));
(3)Return the 3n+1 sequence for n
A Collatz sequence is the sequence where, for a given number n, the next number in the sequence is either n/2 if the number is even or 3n+1 if the number is odd. The sequence always terminates with 1.
So if
n = 13
then
c = [13 40 20 10 5 16 8 4 2 1]
答：
c=n;
while n>1
if mod(n,2)
n =3*n+1;
else
n =n/2;
end
c=[c n];
end