# Single Number

## Question

### Problem Statement

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?

## 题解

「找单数」系列题，技巧性较强，需要灵活运用位运算的特性。根据题意，共有2*n + 1个数，且有且仅有一个数落单，要找出相应的「单数」。鉴于有空间复杂度的要求，不可能使用另外一个数组来保存每个数出现的次数，考虑到异或运算的特性，根据x ^ x = 0x ^ 0 = x可将给定数组的所有数依次异或，最后保留的即为结果。

### C++

class Solution {
public:
/**
* @param A: Array of integers.
* return: The single number.
*/
int singleNumber(vector<int> &A) {
if (A.empty()) {
return -1;
}
int result = 0;

for (vector<int>::iterator iter = A.begin(); iter != A.end(); ++iter) {
result = result ^ *iter;
}

return result;
}
};


### Java

public class Solution {
public int singleNumber(int[] nums) {
if (nums == null || nums.length == 0) return -1;

int result = 0;
for (int num : nums) {
result ^= num;
}

return result;
}
}


### 源码分析

1. 异常处理(OJ上对于空vector的期望结果为0，但个人认为-1更为合理)
2. 初始化返回结果result为0，因为x ^ 0 = x