First Bad Version
@lixiaolun @ChengZhang But if it is exceeding the integer limit it should show some kind of OutOfIndex but why is it showing the
Time Limit Exceed.
@RainbowSecret Ha, I just found I used (start+end)/2 but didn't get overflow. Why? Because I always use uint when I know negative is not possible. This really saves me from a lot of boundary conditions. :)
You don't need a "long long int." Just "long" for low and high.
why if I use
got time exceed limit issue
@lee215 How about a 1-liner using reduce?
my ugly binary search solution :)
Can you please expand on those "detail"?
Do you know the difference between (b-a)/2 + a and (b+a)/2 ? Why this will cause stackoverflow?
You solution is based on there must exist bad version. if all versions are not bad, your solution won't work. Example, n is 1 and 1 is good version.
Did you check if your loop might become infinite loop in some cases?
I get it, thanks
Because your ( Start + end ) value can go beyond Integer.MAX_VALUE and hence giving you a negative value.
Eg : start = 1073741824 , end = 1073741827
if mid = (start + end) /2 the mid = 2147483651/2
Here the numerator is more than the integer range in java. hence making mid = -2147483644/2 = -1073741822 which is wrong
if mid = start + (end - start)/2 then
mid = 1073741824 + 1 = 1073741825
I hope this is helpful! :)
I think it is my experience that tells me to do so ... When we add two INT32, we should always be careful for the overflow.
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Very nice to update res in the loop! You don't really need to use long, simply use int and write mid as mid = start + ((end-start)>>1); would work.
Hmm, I believe I've seen that article before, and thus also that >>> solution, so must have forgotten. I guess I *really* don't care about the issue :-)
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