leetcode-problemset/leetcode-cn/originData/the-masseuse-lcci.json

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            "questionId": "1000023",
            "questionFrontendId": "面试题 17.16",
            "categoryTitle": "LCCI",
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            "title": "The Masseuse LCCI",
            "titleSlug": "the-masseuse-lcci",
            "content": "<p>A popular masseuse receives a sequence of back-to-back appointment requests and is debating which ones to accept. She needs a break between appointments and therefore she cannot accept any adjacent requests. Given a sequence of back-to-back appoint&shy; ment requests, find the optimal (highest total booked minutes) set the masseuse can honor. Return the number of minutes.</p>\r\n\r\n<p><b>Note:&nbsp;</b>This problem is slightly different from the original one in the book.</p>\r\n\r\n<p>&nbsp;</p>\r\n\r\n<p><strong>Example 1: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong> [1,2,3,1]\r\n<strong>Output: </strong> 4\r\n<strong>Explanation: </strong> Accept request 1 and 3, total minutes = 1 + 3 = 4\r\n</pre>\r\n\r\n<p><strong>Example 2: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong> [2,7,9,3,1]\r\n<strong>Output: </strong> 12\r\n<strong>Explanation: </strong> Accept request 1, 3 and 5, total minutes = 2 + 9 + 1 = 12\r\n</pre>\r\n\r\n<p><strong>Example 3: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong> [2,1,4,5,3,1,1,3]\r\n<strong>Output: </strong> 12\r\n<strong>Explanation: </strong> Accept request 1, 3, 5 and 8, total minutes = 2 + 4 + 3 + 3 = 12\r\n</pre>\r\n",
            "translatedTitle": "按摩师",
            "translatedContent": "<p>一个有名的按摩师会收到源源不断的预约请求，每个预约都可以选择接或不接。在每次预约服务之间要有休息时间，因此她不能接受相邻的预约。给定一个预约请求序列，替按摩师找到最优的预约集合（总预约时间最长），返回总的分钟数。</p>\n\n<p><strong>注意：</strong>本题相对原题稍作改动</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong> [1,2,3,1]\n<strong>输出：</strong> 4\n<strong>解释：</strong> 选择 1 号预约和 3 号预约，总时长 = 1 + 3 = 4。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong> [2,7,9,3,1]\n<strong>输出：</strong> 12\n<strong>解释：</strong> 选择 1 号预约、 3 号预约和 5 号预约，总时长 = 2 + 9 + 1 = 12。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong> [2,1,4,5,3,1,1,3]\n<strong>输出：</strong> 12\n<strong>解释：</strong> 选择 1 号预约、 3 号预约、 5 号预约和 8 号预约，总时长 = 2 + 4 + 3 + 3 = 12。\n</pre>\n",
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            "hints": [
                "此题有递归和遍历两种解法，但从递归开始可能更容易一些。",
                "递归解法：每个预约都有两个选择（接受预约或拒绝预约）。作为一种蛮力方法，你可以在所有可能性的地方递归。但是请注意，如果接收了预约请求i，那么你的递归算法应该跳过预约请求i + 1。",
                "递归解法：你可以通过制表法优化这种方法。这种方法的运行时间是多少？",
                "递归解法：记忆法的时间复杂度为O(N)，空间复杂度也为O(N)。",
                "迭代法：对递归法进一步研究。你可以迭代地实现类似的策略吗？",
                "迭代法：从数组的末尾开始，然后向后计算可能是最简单的。",
                "迭代法：注意，你永远不会连续跳过3个预约。为什么不会？因为你总是可以接受中间的预约。",
                "迭代法：如果你选择i，那么将永远不会选择i + 1，但是总会选择i + 2或i + 3。",
                "迭代法：使用一个例子并从后往前计算。你可以很容易地找到子数组{rn}、{rn-1, rn}和{rn-2, ..., rn}。如何使用这些结果快速找到{rn-3, ..., rn}的最优解？",
                "迭代法：如果你预约某一时间段，那就不能预约紧邻的下一时间段，但可以预约之后的任何时间。因此，optimal(ri, ..., rn) = max(ri + optimal(ri+2, ..., rn)，optimal(ri+1, ..., rn))。你可以通过从后往前迭代来解决这个问题。",
                "迭代解法：如果你仔细考虑真正需要的数据，应该能够在O(n)时间复杂度和O(1)额外空间复杂度内解出它。"
            ],
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