題目描述（Task decription）:

A non-empty zero-indexed array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.

For example, in array A such that:

A[0] = 9 A[1] = 3 A[2] = 9
A[3] = 3 A[4] = 9 A[5] = 7
A[6] = 9
the elements at indexes 0 and 2 have value 9,
the elements at indexes 1 and 3 have value 3,
the elements at indexes 4 and 6 have value 9,
the element at index 5 has value 7 and is unpaired.

Write a function:

int solution(vector<int> &A);

that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.

For example, given array A such that:

A[0] = 9 A[1] = 3 A[2] = 9
A[3] = 3 A[4] = 9 A[5] = 7
A[6] = 9
the function should return 7, as explained in the example above.

Assume that:

• N is an odd integer within the range [1..1,000,000];
• each element of array A is an integer within the range [1..1,000,000,000];
• all but one of the values in A occur an even number of times.

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

接替思路：

1. 由題目要求可知，元素個數為奇數個，只有一個元素是未配對的數；
2. 先將數組進行排序（使用泛型算法中的方法）；
3. 遍歷數組，每次前進2個位置，如果第i位置與第i+1位置的元素不相等，則，第i位置的元素就是我們要求的數；

PS：感覺我的這個算法并不是很高級，在此拋磚引玉，期待大神的指點。

參考答案：

// you can use includes, for example:
#include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    sort(A.begin(), A.end());
    
    unsigned i=0;
    for (; i<A.size()-1; i+=2)
    {
        if (A[i] != A[i+1])
        {
            break;
        }
    }
    
    return A[i];
}