Codility lesson 2: Odd occurrenc

Task description
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:

def solution(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.

method 3: 用异或来解决
A xor A = 0
A xor 0 = A
异或List中所有的元素，看最后留下的是谁。（但这是只用于只有一个元素留下来的）

def solution(A):
    # write your code in Python 3.6
    res = 0
    for i in A:
        res = res ^ i
    return res

method 2: Hashing

method1: nested loops

outer loop takes all elements, inner loop counts the times of occurrence, but the time complexity is O(n*n)

def solution(A):
        # write your code in Python 3.6
    m = 0
    l =[]
    
    # write your code in Python 3.6
    for i in range(len(A)):    
        for j in range(len(A)):
            if A[i] == A[len(A)-j-1]:
                m+=1
        if m == 1:
            return A[i]
        m = 0