三次样条插值---三弯矩法（Matlab实现），2024-2024历年华为跳动Python面试真题解析

1.9459101;2.079445;2.1972246;2.3025851];

h=zeros(10,1);

d=zeros(10,1);

u=zeros(10,1);

v=zeros(10,1);

r=zeros(10,1);

l=zeros(10,1);

z=zeros(10,1);

m=zeros(10,1);

for t=1:1:9;

h(t)=x(t+1)-x(t);

end

d(1)=6/h(1)*((y(2)-y(1))/h(1)-1);

d(10)=6/h(9)*(0.1-(y(10)-y(9))/h(9));

for t=1:1:8

u(t+1)=h(t)/(h(t)+h(t+1));

v(t+1)=1-u(t+1);

d(t+1)=6/(h(t)+h(t+1))*((y(t+2)-y(t+1))/(x(t+2)-x(t+1))-(y(t+1)-y (t))/(x(t+1)-x(t)));

end

u(10)=1;v(1)=1;r(1)=d(1);

for t=2:1:10

l(t)=u(t)/r(t-1);

r(t)=d(t)-l(t)*v(t-1);

end

z(1)=d(1);

for t=2:1:10

z(t)=d(t)-l(t)*z(t-1);

end

m(10)=z(10)/r(10);

for t=9👎1

m(t)=(z(t)-v(t)*m(t+1))/r(t);

end

for t=1:1:10

if p>=t&&p<(t+1)

最后

不知道你们用的什么环境，我一般都是用的Python3.6环境和pycharm