### Single Number

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?

### Analysis:

The requirement is O(n) time and O(1) space.

Thus, the "first sort and then find " way is not working.

Also the "hash map" way is not working.

Since we can not sort the array, we shall find a cumulative way, which is not about the ordering.

XOR is a good way, we can use the property that A XOR A = 0, and A XOR B XOR A = B.

So, the code becomes extremely easy.

### Code(C++):

class Solution { public: int singleNumber(int A[], int n) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. int res = A[0]; for (int i=1;i<n;i++){ res = res ^ A[i]; } return res; } };

### Code (Python):

class Solution: # @param A, a list of integer # @return an integer def singleNumber(self, A): res = 0 for a in A: res = res ^ a return res

This comment has been removed by the author.

ReplyDelete