Sequential numbers in a list haskell

Question

I am new to haskell and I was attempting a few coding problems that I previously completed for java, however the following problem has me stumped.

Basically the idea is

Answer 1

This is a solution:

consecutiveOnes :: [Int] -> Bool
consecutiveOnes xs = auxOnes False xs

auxOnes :: Bool -> [Int] -> Bool
auxOnes b [] = False
auxOnes b (x:xs) = case (x==1 && b) of {
    True -> True;
    False -> auxOnes (x==1) xs;
};

Another way would be using the isInfixOf method and asking if [1,1] appears anywhere on your list:

consecutiveOnes :: [Int] -> Bool
consecutiveOnes xs = isInfixOf [1,1] xs

The isInfixOf function takes two lists and returns True iff the first list is contained, wholly and intact, anywhere within the second.

But I'm sure there are many other ways of doing it.

Answer 2

One way would be to use pattern matching to look for consecutive ones at the head of the list, and advance down the list until you find them, or you run out of elements to look at.

consecutiveOnes [] = False
consecutiveOnes (1:1:_) = True
consecutiveOnes (_:xs) = consecutiveOnes xs

Answer 3

You could also do it this way:

consecutiveOnes [] = False
consecutiveOnes xs = any (== (1,1)) $ zip xs (tail xs)

If

 xs == [0,1,1,2,3,4]

Then

tail xs == [1,1,2,3,4]

Zipping them together you get a list of pairs, where each pair is an element of the list and the element after it.

zip xs (tail xs) == [(0,1),(1,1),(1,2),(2,3),(3,4)]