function val=longrun(a)
a_len=length(a); %计算a的长度
c=ones(a_len,1); %初始化对应位置的连续次数
for num_a=1:(a_len-1)
n_val=1;
for next_a=(num_a+1):a_len
if a(num_a)==a(next_a) %如果这个数字和后面的数字相等，则n_val+1，并进行下一次循环
n_val=n_val+1;
c(num_a)=n_val;
else
c(num_a)=n_val; %如果这个数字和后面数字不等，则把当前的n_val赋值给对应数字的连续次数，并跳出内层for循环
break
end
end
end
a_conti=max(c); %找出c数组中最大的连续次数
max_location=find(c==a_conti); %对应最大连续次数的位置
val=a(max_location); %找出对应在a中的位置
end

Test 1 and 4 are wrong

find(vertcat(1,diff(a(:)),1));
val=a(ans(diff(ans)==max(diff(ans))))

@Piero Cimule
Could you please comment on this method? It is very unlikely for me as a beginner in Matlab to have such a concise code. You are more than welcome to elaborate here or via my e-mail address zasevzasev42@gmail.com

Could you check this for me, please?
[ra,ca]=size(a);
if ra>ca
a=a';
end

j=1;
V=[a(1,1);1];
for i=2:size(a,2)
X=a(1,i);
if V(1,j)==X
V(2,j)=V(2,j)+1;
else
j=j+1;
V(1,j)=X;
V(2,j)=V(2,j)+1;
end

end

val=V(V(2,:)==max(V(2,:)))
if ra>ca
val=val';
end

How else could you solve in any other traditional programming language?

It was not so easy...

i'd really like to see a better solution that mine, one that uses difference equations..mine is size 129 and is just a control loop. Guess i've got to search in solution map.

These are non-constructive solutions and should be removed.

function ans=longrun(a)
d = find([true;diff(a(:))~=0;true]);
f = [d(1:end-1) diff(d)];
a(f(f(:,2)==max(f(:,2))))
end

very clever!