[LeetCode] - Single Number

Description: Given an array of integers, every element appears twice except for one. Find that single one.

Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

这是整个LeetCode上面做的第一道题。没啥好说的，XOR即可。

public class Solution {
public int singleNumber(int[] A) {
int single = 0;
for(int i=0; i < A.length; i++) {
single ^= A[i];
}
return single;
}
}

http://stackoverflow.com/questions/1089987/given-an-array-of-numbers-except-for-one-number-all-the-others-occur-twice-gi

Assuming you can 

XOR

 the
numbers, that is the key here, I believe, because of the following properties:

XOR

 is
commutative and associative (so the order in which it's done is irrelevant).

a number 

XOR

ed
with itself will always be zero.

zero 

XOR

ed
with a number will be that number.

So, if you simply 

XOR

 all
the values together, all of the ones that occur twice will cancel each other out (giving 0) and the one remaining number (

n

)
will 

XOR

 with
that result (0) to give 

n

.