leetcode-problemset/leetcode-cn/originData/intersection-of-multiple-arrays.json

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        "question": {
            "questionId": "2331",
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            "categoryTitle": "Algorithms",
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            "title": "Intersection of Multiple Arrays",
            "titleSlug": "intersection-of-multiple-arrays",
            "content": "Given a 2D integer array <code>nums</code> where <code>nums[i]</code> is a non-empty array of <strong>distinct</strong> positive integers, return <em>the list of integers that are present in <strong>each array</strong> of</em> <code>nums</code><em> sorted in <strong>ascending order</strong></em>.\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [[<u><strong>3</strong></u>,1,2,<u><strong>4</strong></u>,5],[1,2,<u><strong>3</strong></u>,<u><strong>4</strong></u>],[<u><strong>3</strong></u>,<u><strong>4</strong></u>,5,6]]\n<strong>Output:</strong> [3,4]\n<strong>Explanation:</strong> \nThe only integers present in each of nums[0] = [<u><strong>3</strong></u>,1,2,<u><strong>4</strong></u>,5], nums[1] = [1,2,<u><strong>3</strong></u>,<u><strong>4</strong></u>], and nums[2] = [<u><strong>3</strong></u>,<u><strong>4</strong></u>,5,6] are 3 and 4, so we return [3,4].</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [[1,2,3],[4,5,6]]\n<strong>Output:</strong> []\n<strong>Explanation:</strong> \nThere does not exist any integer present both in nums[0] and nums[1], so we return an empty list [].\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= sum(nums[i].length) &lt;= 1000</code></li>\n\t<li><code>1 &lt;= nums[i][j] &lt;= 1000</code></li>\n\t<li>All the values of <code>nums[i]</code> are <strong>unique</strong>.</li>\n</ul>\n",
            "translatedTitle": "多个数组求交集",
            "translatedContent": "<p>给你一个二维整数数组 <code>nums</code> ，其中 <code>nums[i]</code> 是由 <strong>不同</strong> 正整数组成的一个非空数组，按 <strong>升序排列</strong> 返回一个数组，数组中的每个元素在 <code>nums</code>&nbsp;<strong>所有数组</strong> 中都出现过。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [[<em><strong>3</strong></em>,1,2,<em><strong>4</strong></em>,5],[1,2,<em><strong>3</strong></em>,<em><strong>4</strong></em>],[<em><strong>3</strong></em>,<em><strong>4</strong></em>,5,6]]\n<strong>输出：</strong>[3,4]\n<strong>解释：</strong>\nnums[0] = [<em><strong>3</strong></em>,1,2,<em><strong>4</strong></em>,5]，nums[1] = [1,2,<em><strong>3</strong></em>,<em><strong>4</strong></em>]，nums[2] = [<em><strong>3</strong></em>,<em><strong>4</strong></em>,5,6]，在 nums 中每个数组中都出现的数字是 3 和 4 ，所以返回 [3,4] 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [[1,2,3],[4,5,6]]\n<strong>输出：</strong>[]\n<strong>解释：</strong>\n不存在同时出现在 nums[0] 和 nums[1] 的整数，所以返回一个空列表 [] 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= sum(nums[i].length) &lt;= 1000</code></li>\n\t<li><code>1 &lt;= nums[i][j] &lt;= 1000</code></li>\n\t<li><code>nums[i]</code> 中的所有值 <strong>互不相同</strong></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<int> intersection(vector<vector<int>>& nums) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def intersection(self, nums):\n        \"\"\"\n        :type nums: List[List[int]]\n        :rtype: List[int]\n        \"\"\"",
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                "Keep a count of the number of times each integer occurs in nums.",
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