[LeetCode OJ] Single Number之一 ——Given an array of integers, every element appears twice except for one. Find that single one.

 1 class Solution {

 2 public:

 3     int singleNumber(int A[], int n) {

 4         int i,j;

 5         for(i=0; i<n; i++)

 6         {

 7             for(j=i+1; j<n; j++)

 8             {

 9                 if(A[j]==A[i])

10                 {

11                     int temp = A[i+1];

12                     A[i+1] = A[j];

13                     A[j] = temp;

14                     i++;

15                     break;

16                 }

17             }

18             if(j==n)

19                 return A[i];

20         }

21     }

22 };

上面的是传统方法，时间复杂度是O(n²)，空间复杂度是O(1)。

1 class Solution {

2 public:

3     int singleNumber(int A[], int n) {

4         int result=0;

5         for(int i=0; i<n; i++)

6             result = result ^ A[i];

7         return result;

8     }

9 };

用位异或运算实现，时间复杂度和空间复杂度均为O(1).

运用了异或运算具有交换律和结合律的性质。

交换律： a^b = b^a

结合律： (a^b)^c = a^(b^c)

另外，a^a=0。

[LeetCode OJ] Single Number之一 ——Given an array of integers,
every element appears twice except for one. Find that single one.