问题
I'm trying to solve the following puzzle:
Given a stream of numbers (only 1 iteration over them is allowed) in which all numbers appear 3 times, but 1 number appear only 2 times, find this number, using O(1) memory.
I started with the idea that, if all numbers appeared 2 times, and 1 number only once, I could use xor operation between all numbers and the result would be the incognito number.
So I want to extend this idea to solve the puzzle. All I need is a xor-like function (or operator), which would yield 0 on the third apply:
SEED xor3 X xor3 X xor3 X = SEED
X xor3 Y xor3 SEED xor3 X xor3 Y xor3 Y xor3 X = SEED
Any ideas for such a function?
回答1:
Regard XOR as summation on each bit of a number expressed in binary (i.e. a radix of 2), modulo 2.
Now consider a numerical system consisting of tribits 0, 1, and 2. That is, it has a radix of 3.
The operator T now becomes an operation on any number, decomposed into this radix. As in XOR, you sum the bits, but the difference is that operator T is ran in modulo 3.
You can easily show that a T a T a is zero for any a. You can also show that T is both commutative and associative. That is necessary since, in general, your sequence will have the numbers jumbled up.
Now apply this to your list of numbers. At the end of the operation, the output will be b where b = o T o and o is the number that occurs exactly twice.
回答2:
Your solution for the simpler case (all number appear twice, one number appears once) works since xor operates on each bit x as
x xor x = 0 and 0 xor x = x
xor is basically a bit-wise summation modulus 2. You would need the base-3 equivalent: Transform each number into a base-3 representation. And then use summation modulus 3 for each decimal:
0 1 2
0 0 1 2
1 1 2 0
2 2 0 1
Call this operation xor3. Now you have for each decimal x:
x xor3 x xor3 x = 0 and 0 xor3 x = x
If you apply that to all your numbers then all values that appear 3 times will vanish. The result is x xor3 x of the number x that appears twice. You need to apply decimal-wise division by 2 modulus 3.
I believe there are more efficient ways to implement that. The advantage of the xor function in the first case relies on the fact that xor is a natural base-2 operation. Is there any practical application for that?
回答3:
This approach is a bit fragile: If the precondition (all numbers appear 3 times except one that appears twice) breaks the algorithm will not help you.
Take a Map with int-keys and int-values. Then walk through your numbers and for each number x increase each the according value. If x is a new key take 0 as start value.
Then you can analyze it easily: Walk through all keys and check the cardinality. It should be three for all keys except one that should be two. This is more robust and my gut feeling says it is also faster.
来源：https://stackoverflow.com/questions/23887820/three-way-xor-like-function