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]