% If you are only looking for an answer to this scenario the easiest way is
% as follows:

A = [0 0 1 1];
B = [0 0 1 1];
nn = length(A);

keepset = [0 0 1 1;0 1 0 1];
keepset = keepset(:,randperm(nn))

% If you want a more general solution for arbitrary A & B (for example)
A = [0 0 0 1 1 1 2 2 2];
B = [0 0 0 1 1 1 2 2 2];
nn = length(A);

Ai = A(randperm(nn));
Bi = B(randperm(nn));

% initialize keepset with the first combination of A & B
keepset = [Ai(1);Bi(1)];

loopcnt = 0;
while (size(keepset,2) < nn) 

    % randomize the elements in A and B independently
    Ai = A(randperm(nn));
    Bi = B(randperm(nn));
    % test each combination of Ai and Bi to see if it is already in the
    % keepset
    for ii = 1:nn
        tstcombo = [Ai(ii);Bi(ii)];
        matchtest = bsxfun(@eq,tstcombo,keepset);
        matchind = find((matchtest(1,:) & matchtest(2,:)));
        if isempty(matchind)
            keepset = [keepset tstcombo];
        end
    end
    loopcnt = loopcnt + 1;
    
    if loopcnt > 1000
        disp('Execution halted after 1000 attempts')
        break
    elseif (size(keepset,2) >= nn)
        disp(sprintf('Completed in %0.f iterations',loopcnt))
    end
        
end
keepset

下面是一个可能的解决方案：

A = [0 0 1 1];
B = [0 0 1 1];

% Randomize A and B independently
ARand = randperm(length(A));
A = A(ARand);
BRand = randperm(length(B));
B = B(BRand);

% Keep randomizing A and B until the condition is met
while sum(A+B) ~= 1 && sum(A+B) ~= length(A)
    ARand = randperm(length(A));
    A = A(ARand);
    BRand = randperm(length(B));
    B = B(BRand);
end

此解决方案检查A+B中的元素之和是否为1或A的长度，这表示A+B中只有一个元素分别为0或2。如果不满足这两个条件中的任何一个，则再次随机化向量A和B。