Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.
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Example 1:
\n\n\nInput: nums1 = [1,2,2,1], nums2 = [2,2]\nOutput: [2,2]\n\n\n
Example 2:
\n\n\nInput: nums1 = [4,9,5], nums2 = [9,4,9,8,4]\nOutput: [4,9]\nExplanation: [9,4] is also accepted.\n\n\n
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Constraints:
\n\n1 <= nums1.length, nums2.length <= 10000 <= nums1[i], nums2[i] <= 1000\n
Follow up:
\n\nnums1's size is small compared to nums2's size? Which algorithm is better?nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?给你两个整数数组 nums1 和 nums2 ，请你以数组形式返回两数组的交集。返回结果中每个元素出现的次数，应与元素在两个数组中都出现的次数一致（如果出现次数不一致，则考虑取较小值）。可以不考虑输出结果的顺序。
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示例 1：
\n\n\n输入：nums1 = [1,2,2,1], nums2 = [2,2]\n输出：[2,2]\n\n\n
示例 2:
\n\n\n输入：nums1 = [4,9,5], nums2 = [9,4,9,8,4]\n输出：[4,9]\n\n
\n\n
提示：
\n\n1 <= nums1.length, nums2.length <= 10000 <= nums1[i], nums2[i] <= 1000\n\n
进阶：
\n\nnums1 的大小比 nums2 小，哪种方法更优？nums2 的元素存储在磁盘上，内存是有限的，并且你不能一次加载所有的元素到内存中，你该怎么办？\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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