use dynamic programming
seq:1 2 7 3 2 8 4 8 5
dp :1 2 3 4 2 3 4 3 4
dp[i] -> dp[i+1]
or
dp[i] -> dp[j] while seq[i] == seq[j]
Notice that if much j exists, just goto the smallest j
the algorithm is O(N)

P.S : why netflow doesn't work now ? :)