问题15.在二进制序列中找到1的最长序列。

noice

cool

function y = lengthOnes(x)
y = double(x == '1');
l = numel（y）;
a = tril（一个（l，l））;
C = mat2cell(A,ones(1,L),L);
D = cellfun(@(c) conv(c,y),C,"UniformOutput",false);
e = cellfun（@（c）sum（c），c）;
f = cellfun（@（c）max（c），d）;
r = max（e（f == e））;
if(isempty(R))
y = 0;
else
y = r;
结尾
结尾

function y = lengthOnes(x)
cnt=0;
a = [];
l = strlength（x）;
for i=1:L
if x(i)=='1'
cnt=cnt+1;
else
cnt=0;
结尾
a（i）= cnt;
结尾
[y,index]=max(a);
结尾

Excellent work man :)

They are characters not doubles!

function y = lengthOnes(x)
a = [];
idx = [1,find(x=='0') + 1,length(x)+2];
le = length(idx);
a（1：le-1）= idx（2：le）-idx（1：le-1）-1;
y = max（a）;
结尾

实际上不需要言论

Cool

解决d on iPhone

@Martin在诺基亚X2-02上解决了

Any suggestions on how to improve more ?

如何改进代码？

a = regexp（x，'0+'，'split'）
a=a{max(length(a))}
y =长度（a）
With this code, every test passes except 4th one. Why is it so?

有什么方法可以改善该算法？
when y is empty vector, max(y) is not returning zero which is making my code complicated

它在我的MATLAB程序中运行良好。

您的MATLAB需要多少时间？科迪有50秒的限制

Aditya Jain
Do you have a source for that time limit info?

probably the most inefficient code ever written :)

我喜欢这个

efficient idea

凉爽的！

I know it's not very efficient, but I sort of liked an idea behind this loop. Hence shared.

Great solution that's easy to understand adn translatable to future use.

非常不熟悉“ Regexp”功能会导致非常愚蠢的代码...

hmmmmm.... ;-) nice

动态表达！这太酷了，我不知道Matlab有这个，您可以编写一个很难阅读的MATLAB代码。为什么49？

Honestly, right now I find it hard to read as well. 49 is the decimal representation of the ASCII character '1' (type +'1' in the Matlab Command Window).

最后，在科迪中正确使用regexp

您能解释您的代码吗？