2010/12/5 juver++ <avpostni...@gmail.com>:
> Wrong.
> Counterexample: 1 2 2 2 2 6
suppose you mean 1 3 3 3 3 6?
1) divide-conquer
2) search binary, in each sub array [s, t],
mid = (s+t)/2
if t < array[mid]:
// no need to check the part after mid, all will fail
else if s > array[mid]:
// no need to check the part before mid, all will fail
// if we have the elements unique
if mid < array[mid]:
// no need to check the part after mid, all will fail
else if mid > array[mid]:
// no need to check the part before mid, all will fail
3) when the elements are unique, in [s, t],
if (array[s] = s && array[t] = t)
// all in [s, t] is ok
but I think its complexity will be O(n) in the worse case if elements
could be equal.
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