题目如下：

You are given an integer array nums and an integer k.

In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array.

Return the maximum number of operations you can perform on the array.

Example 1:

Input: nums = [1,2,3,4], k = 5
Output: 2
Explanation: Starting with nums = [1,2,3,4]:
- Remove numbers 1 and 4, then nums = [2,3]
- Remove numbers 2 and 3, then nums = []
There are no more pairs that sum up to 5, hence a total of 2 operations.

Example 2:

Input: nums = [3,1,3,4,3], k = 6
Output: 1
Explanation: Starting with nums = [3,1,3,4,3]:
- Remove the first two 3's, then nums = [1,4,3]
There are no more pairs that sum up to 6, hence a total of 1 operation.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109
• 1 <= k <= 109

解题思路：首先统计出每个num出现的次数，然后求num和k-num出现次数的较小值，即为这对组合可操作的次数，注意要考虑 num = k- num的情况。

代码如下：

class Solution(object):
    def maxOperations(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        dic = {}
        res = 0
        for num in nums:
            dic[num] = dic.setdefault(num,0) + 1
        for key in dic.iterkeys():
            if dic[key] <= 0:continue
            elif k - key == key:
                res += dic[key]/2
                dic[key] = dic[key] % 2
            elif k - key in dic:
                pair = min(dic[key],dic[k-key])
                res += pair
                dic[key] -= pair
                dic[k-key] -= pair
        return res