I know a solution based on hashing and dynamic programming. Let f(x) be the hash function. The trick is the hash-table value. Consider the longest interval contained in the list, which either starts or ends with x. Then h[f(x)] = y, where y is the other end of that interval. Note that length of that interval will be abs( x - y ) + 1. The algorithm description will make clear why to store that value.
Move over the list. Let i be current index, x := list[ i ] - current number. Now
1. if h[f(x)] is not empty, then we've met number x before. Nothing to do, continue.
2. Check h[f(x-1)] and h[f(x+1)].
2.1. If they're both not empty, that means we've already met x-1 and x+1, and we know some intervals [a..x-1] and [x+1..b] which we've already met in the list. We know it because a=h[f(x-1)] and b=h[f(x+1)] by definition of h. Now when we got x, it means that we now have met the whole interval [a,b], so we update values as follows: h[f(a)] := b and h[f(b)] := a.
Also set h[f(x)] to some value (let's say x, not to impact the answer), just so that next time we meet x in the list, we ignore it. x has already done his job.
2.2. If only one of them is set, let's say h[f(x-1)] = a, that means we've already met some interval [a..x-1], and now it's extended with x. Update will be h[f(a)] := x and h[f(x)] := a.
2.3. If none of them is set, that means we've met neither x-1, nor x+1, and the biggest interval containing x we've already met is the single [x] itself. So set h[f(x)] := x.
Finally, to get the answer, pass over the whole list and take maximum abs( x - h[f(x)] ) + 1 for all x.
