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金字塔矩阵练习

2011-08-16 13:15:02| 分类： M&M | 标签： |举报 |字号大中小订阅

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要求是编写一个M函数，接受一个奇数，然后产生一个N*(2N-1)的矩阵，形式如下：

0 0 0 0 1 0 0 0 0

0 0 0 2 1 2 0 0 0

0 0 3 2 1 2 3 0 0

0 4 3 2 1 2 3 4 0

5 4 3 2 1 2 3 4 5

想了一下，有了个结果：

function A=pyramid(n)

if mod(n,2)==0

error('n should be an odd number!');

end

A=zeros(n,2*n-1);

for i=1:n

A(n-i+1:n,i)=n-i+1;

A(n-i+1:n,2*n-i)=n-i+1;

end

测试一下：

>> pyramid(7)

ans =

0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 2 1 2 0 0 0 0 0

0 0 0 0 3 2 1 2 3 0 0 0 0

0 0 0 4 3 2 1 2 3 4 0 0 0

0 0 5 4 3 2 1 2 3 4 5 0 0

0 6 5 4 3 2 1 2 3 4 5 6 0

7 6 5 4 3 2 1 2 3 4 5 6 7

应该是没有什么问题了。这个练习中学到的技巧是避免了使用双重循环，而是使用的这种形式：A(n-i+1:n,i)=n-i+1，直接操作n-i+1到n行的数据。

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