XOR to find Duplicates in an array

问题

I have seen the solution for this problem in this thread -> How to find a duplicate element in an array of shuffled consecutive integers?

But the problem I am now having is little varied from it.

int arr[10] = {1,2,3,4,5,6,7,8,4,9};
    int a= 0;
    for(int i=0;i<10;i++) {
     a= a^ arr[i] ^i;
    }
    cout<<a;

Consider the above mentioned code snippet. Things work fine as it is. But when I add a 0 to the above mentioned array like, int arr[11] = {0,1,2,3,4,5,6,7,8,4,9}; I am not getting the proper duplicate element. Can somebody correct me the mistake I am making here?

回答1:

The trick relies on the values being between 1 and n. If the numbers are in some other range you'll have to offset them.

static const int n = 11;
int arr[n] = {0,1,2,3,4,5,6,7,8,4,9};
int offset = 1;
int a= 0;
for(int i=0;i<n;i++) {
 a= a^ (arr[i]+offset) ^i;
}
cout<< (a-offset);

回答2:

I'm guessing it could have to do with the fact that the i value would then be the same as arr[i]

thats like doing:

00000001 ^ 00000001

which equals 00000000

and I may be thinking incorrectly but wouldn't that screw up the process?

回答3:

I think i see what happened. you probably changed the loop to be

for(int i=0; i<11; i++)
  ...

since you added an extra element to the loop. the problem is that it's not XORing with 10, which is not a number in your array. So, the algorithm stops working.

int arr[11] = {0,1,2,3,4,5,6,7,8,4,9};
int a= 0;
for(int i=0;i<10;i++) {
  a= a^ arr[i] ^i;
}
a = a^arr[10];

cout<<a;

now, its XORing nums from 0 to 9 twice (which equals zero) and 4 an additional time, so the result should be 4.

来源：https://stackoverflow.com/questions/8018086/xor-to-find-duplicates-in-an-array

标签

c++

algorithm

xor