这是我试图找到连续零的代码,大约为5或更多.
a=[0,0,0,0,0,0,0,0,9,8,5,6,0,0,0,0,0,0,3,4,6,8,0,0,9,8,4,0,0,7,8,9,5,0,0,0,0,0,8,9,0,5,8,7,0,0,0,0,0];
[x,y]=size(a);
for i=0:y
i+1;
k=1;
l=0;
n=i;
count=0;
while (a==0)
count+1;
break;
n+1;
end
if(count>=5)
v([]);
for l=k:l<n
v(m)=l+1;
m+1;
end
end
count=1;
i=n;
end
for i = o : i<m
i+1;
fprintf('index of continous zero more than 5 or equal=%d',v(i));
end
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如果要查找运行n或更多零的起始索引:
v = find(conv(double(a==0),ones(1,n),'valid')==n); %// find n zeros
v = v([true diff(v)>n]); %// remove similar indices, indicating n+1, n+2... zeros
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在你的例子中,这给出了
v =
1 13 34 45
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