leetcode-problemset/leetcode-cn/originData/living-people-lcci.json

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        "question": {
            "questionId": "100351",
            "questionFrontendId": "面试题 16.10",
            "categoryTitle": "LCCI",
            "boundTopicId": 91325,
            "title": "Living People LCCI",
            "titleSlug": "living-people-lcci",
            "content": "<p>Given a list of people with their birth and death years, implement a method to compute the year with the most number of people alive. You may assume that all people were born between 1900 and 2000 (inclusive). If a person was alive during any portion of that year, they should be included in that year&#39;s count. For example, Person (birth= 1908, death= 1909) is included in the counts for both 1908 and 1909.</p>\r\n\r\n<p>If there are more than one years&nbsp;that have the most number of people alive, return the smallest one.</p>\r\n\r\n<p>&nbsp;</p>\r\n\r\n<p><strong>Example: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>\r\nbirth = [1900, 1901, 1950]\r\ndeath = [1948, 1951, 2000]\r\n<strong>Output: </strong> 1901\r\n</pre>\r\n\r\n<p>&nbsp;</p>\r\n\r\n<p><strong>Note: </strong></p>\r\n\r\n<ul>\r\n\t<li><code>0 &lt; birth.length == death.length &lt;= 10000</code></li>\r\n\t<li><code>birth[i] &lt;= death[i]</code></li>\r\n</ul>\r\n",
            "translatedTitle": "生存人数",
            "translatedContent": "<p>给定 N 个人的出生年份和死亡年份，第 <code>i</code> 个人的出生年份为 <code>birth[i]</code>，死亡年份为 <code>death[i]</code>，实现一个方法以计算生存人数最多的年份。</p>\n\n<p>你可以假设所有人都出生于 1900 年至 2000 年（含 1900 和 2000 ）之间。如果一个人在某一年的任意时期处于生存状态，那么他应该被纳入那一年的统计中。例如，生于 1908 年、死于 1909 年的人应当被列入 1908 年和 1909 年的计数。</p>\n\n<p>如果有多个年份生存人数相同且均为最大值，输出其中最小的年份。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nbirth = [1900, 1901, 1950]\ndeath = [1948, 1951, 2000]\n<strong>输出：</strong> 1901\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt; birth.length == death.length &lt;= 10000</code></li>\n\t<li><code>birth[i] &lt;= death[i]</code></li>\n</ul>\n",
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                {
                    "name": "Array",
                    "slug": "array",
                    "translatedName": "数组",
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                    "name": "Counting",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    int maxAliveYear(vector<int>& birth, vector<int>& death) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def maxAliveYear(self, birth, death):\n        \"\"\"\n        :type birth: List[int]\n        :type death: List[int]\n        :rtype: int\n        \"\"\"",
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                    "code": "public class Solution {\n    public int MaxAliveYear(int[] birth, int[] death) {\n\n    }\n}",
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                    "code": "/**\n * @param {number[]} birth\n * @param {number[]} death\n * @return {number}\n */\nvar maxAliveYear = function(birth, death) {\n\n};",
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                    "code": "# @param {Integer[]} birth\n# @param {Integer[]} death\n# @return {Integer}\ndef max_alive_year(birth, death)\n\nend",
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                    "code": "(define/contract (max-alive-year birth death)\n  (-> (listof exact-integer?) (listof exact-integer?) exact-integer?)\n\n  )",
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            "hints": [
                "方案 1：你能计算出每年有多少人活着吗？",
                "方案1：用散列表或数组试试，将出生年份映射到该年还有多少人活着。",
                "解法2：如果对年份排序会如何？你会根据什么排序？",
                "解法2：你真的有必要匹配出生年份和死亡年份吗？当一个特定的人死了，会有什么关系，或者你只是需要一份死亡年份的清单？",
                "解法2：观察到人是“可替代的”，不管谁出生，何时死亡。你需要的只是一份出生年份和死亡年份的列表。这可能会使你对人员列表的排序变得更加容易。",
                "解法2：尝试创建一份排序的出生列表和一份排序的死亡列表。通过遍历两个列表，你能追踪任意时间活着的人的数量吗？",
                "每个出生增加一个人，每个死亡移除一个人。尝试编写一份人员列表（出生年份和死亡年份）示例，然后将其重新格式化为每年的列表，出生时加1，死亡时减1。",
                "解法3：如果你创建了一个年份数组并保存每个年份的人口变化会如何？你能找到人口最多的那一年吗?",
                "解法3：注意这个问题中的细节。你的算法/代码是否考虑一个在出生的同一年去世的人？这个人应该被计算为人口总数中的一人。"
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            "sampleTestCase": "[1900,1901,1950]\n[1948,1951,2000]",
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