问题
I'm learning algorithms/big o and i was just curious about this.
The use of
mid = (low+high)/2;
in order to get the midpoint for a binary search algorithm is generally discouraged because of the possibility of an overflow error. Why would this cause an overflow error to occur, and how does
mid = low + (high-low)/2;
prevent this error?
Thanks.
回答1:
In the first case you calculate the value (low+high) which might be too huge to fit into an int, if low and high are both huge enough (say if both are equal to 2^30+1 /or even greater/). In the second case, you don't calculate (low+high), you do a small trick and you go through the expression (high-low) and that expression is much safer with respect to int overflow.
Still, if you don't have an array whose size is greater than 2^30 (which is quite a huge array anyway), I don't see how you can run into an int overflow even when using the first expression. So I would just use the first one in most cases and not worry.
回答2:
Lets assume the worst situation for integer. We are sure that mid = (low + high)/2 can never give a overflow if we look at the bigger picture. Assuming that low = 2ˆ31 - 1 and high = 2ˆ31 - 1 ( the highest integer values) the total calculation will be [2*(2ˆ31-1)] / 2 = 2ˆ31 - 1. ( the biggest number int can hold)
However the way we calculate it matters.
If it is calculated using (low + high)/2 it will simply give an overflow in our situation because first it will try to sum low( 2ˆ31 -1 ) and high ( 2ˆ31 -1 ) and BAM. 2*(2ˆ31-1) is an overflow. It can not divide it by two because it can not store this value initially in an int.
If it is calculated using low + (high-low)/2 it can not give an overflow . Think it that way, low + (high - low)/2 = (low + x) where x = (high - low)/2. In order to create an overflow we should pick big (in our situation biggest for int) numbers for both sides. However if we pick low the biggest, then x will be the smallest( because we are subtracting low from high in x) and vice versa.
回答3:
Each primitive datatype has a maximum and minimum value. When you exceed the maximum or minimum value for that datatype it wraps around to the min or max respectively.
mid = (low+high)/2;
In this case if low+high>MAX_VALUE then you will get strange results.
Heres a fun test you can try out
int high = Integer.MAX_VALUE;
int low = 3*Integer.MAX_VALUE/4;
int mid = (high+low)/2;
System.out.println(mid); // -805306369
By changing your math you can remove the chance that you will execde that maximum or minimum value by subtracting them instead of adding them.
来源：https://stackoverflow.com/questions/24317360/midpoint-formula-overflow-error