코딜리티 - OddOccurrencesInArray(c++)

문제

A non-empty 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.

Write an efficient algorithm for the following assumptions:

• 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.

풀이

배열을 정렬 하면 같은 값을 가지는 원소끼리 짝지어 지게 된다.

배열을 처음부터 순회하면서 붙어 있는 두개의 원소가 같은 원소라면 인덱스를 2증가시켜 다음 원소로 넘어가고

붙어 있는 두개의 원소가 다른 원소라면 해당 원소는 하나만 존재하는 원소이므로 answer에 담고 while문을 종료한다.

#include <algorithm>

int solution(vector<int> &A) {
    int answer = 0;
    
    sort(A.begin(), A.end());
    
    unsigned int i = 0;
    while(i<A.size()){
        if(i+1 != A.size() - 1 && A[i] == A[i+1]) i+=2;
        else{
            answer = A[i];
            break;
        }
    }
   
    return answer;
}