1、純遞歸，巨慢

function F = fibo01(n)

% ? The recursion solution of the fibonacci sequence

% ? call the fibo(n) , n is a positive integer number

arguments

? ?n (1,1) {mustBeInteger,mustBePositive}

end

if n <= 2

? ?F = 1;

else

? ?F = fibo01(n-1)+fibo01(n-2);

end

end

2、加入備忘錄，自底向上，預分配內(nèi)存，空間換時間，快多了

function ?F = fibo02(n)

%The memo and End-Top Method of the fibonacci sequences

% ? call fibo02(n) to get the ans, n is a positive integer

arguments

? ?n (1,1) {mustBeInteger,mustBePositive}

end

memo = sym(ones(1,n)); ?% preallocate the memory and tradeoff space with time

if n <= 2

? ?F = memo(n);

else

? ?for i = 3:n

? ? ? ?memo(i) = memo(i-1)+ memo(i-2);

? ?end

? ?F = memo(n);

end

3、加入備忘錄，自頂向上，還是遞歸

function F = fibo03(n)

%the memo version of fibonacci sequences

% ? call fibo03(n) ,here n is a positive ?integer number

arguments

? ?n (1,1) {mustBeInteger,mustBePositive}

end

global memo

memo = sym(ones(1,n));

F = fibo_03(n);

end

function F = fibo_03(n)

%the memo version of fibonacci sequences

% ? call fibo03(n) ,here n is a positive ?integer number

arguments

? ?n (1,1) {mustBeInteger,mustBePositive}

end

global memo

if n<=2

? ?F = memo(n);

elseif memo(n) == 1

? ?memo(n) = fibo_03(n-1)+fibo_03(n-2);

? ?F = memo(n);

else

? ?F = memo(n);

end

end

4、雙指針，省時省空間

function F = fibo04(n)

%double pointer solution of fibonacci sequences

% ? call fibo04(n),n is a positive integer number here

arguments

? ?n (1,1) {mustBeInteger,mustBePositive}

end

a = sym(1); % initialize the two pointer

b = sym(1);

if n<=2

? ?F = 1;

else

? ?for i = 3:n

? ? ? ?c = a+b; ?% solve the i-th fabonacci

? ? ? ?a = b; ? ?% swap the a,b,c ,prepare for next interation

? ? ? ?b = c;

? ?end

? ?F = c;

end

很明顯，雙指針的速度要快很多，其次是向上模擬，然后是帶備忘錄的遞歸，最次是遞歸，根本算不出結果，由于使用了一些符號化以及輸入?yún)?shù)的限定，提高了魯棒性，同時也附加了額外的開銷