leetcode350之实现求解两数组交集（包含重复元素）

给定两个数组，编写一个函数来计算它们的交集。

说明：

• 输出结果中每个元素出现的次数，应与元素在两个数组中出现的次数一致。
• 我们可以不考虑输出结果的顺序

 def binarySearch(nums, target):
     '''
     在数组中二分查找指定元素
     :param nums:
     :param target:
     :return:
     '''
     left, right = 0, len(nums) - 1
     while left <= right:
         mid = left + (right - left) // 2
         if nums[mid] == target:
             return True
         elif nums[mid] < target:
             left = mid + 1
         else:
             right = mid - 1

     return False

 def intersect1(num1, num2):
     '''
     求两个列表交集元素
     :param num1:
     :param num2:
     :return:
     '''
     array1 = []
     num1 = sorted(num1)
     for i in num2:
         if binarySearch(num1, i):
             array1.append(i)
             num1.remove(i)
     array1 = sorted(array1)

     return array1

 print("==========测试intersect()==========")
 nums1 = [1, 3, 1, 2, 4, 2]
 nums2 = [4, 8, 1, 2, 1, 3, 8, 2]
 array1 = intersect1(nums1, nums2)
 print("array1=", array1)

 def intersect2(nums1, nums2):
     '''

     :param nums1:
     :param nums2:
     :return:
     '''
     array2 = []
     temp_array2 = set(nums1) & set(nums2)  # temp_array2 is a set
     for i in temp_array2:
         array2 += [i] * min(nums1.count(i), nums2.count(i))
     return array2

 print("+++++++++++测试intersect2()+++++++++++++")
 array2 = intersect2(nums1, nums2)
 print("array2=", array2)

 def intersect3(nums1, nums2):
     '''

     :param nums1:
     :param nums2:
     :return:
     '''
     array3 = []
     for i in nums1:
         if i in nums2:
             array3.append(i)
             nums2.remove(i)
     array3 = sorted(array3)
     return array3

 print("----------------测试intersect3()-----------")
 array3 = intersect3(nums2, nums1)
 print("array3=", array3)

输出：

==========测试intersect()==========
array1= [1, 1, 2, 2, 3, 4]
+++++++++++测试intersect2()+++++++++++++
array2= [1, 1, 2, 2, 3, 4]
----------------测试intersect3()-----------
array3= [1, 1, 2, 2, 3, 4]