python经典算法题之三:OddOccurrencesInArray

本文介绍了一种算法,用于从包含奇数个整数的数组中找出唯一未配对的元素。该算法适用于数组中除一个元素外所有元素均成对出现的情况。通过遍历数组并使用字典记录每个元素出现的次数,最终找到出现次数为奇数的元素。

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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:

class Solution { public int solution(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).

Solution:

def solution(p):
    d = {}
    result = []
    for i in p:
        if i not in d.keys():
            d[i] = 1
        else :
            d[i] += 1
    for i,j in d.items():
        if j%2 == 0:
            pass
        else:
            result.append(i)
    for index, item in enumerate(result):
        return int(item)

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