update

leetcode-cn/originData/grumpy-bookstore-owner.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>有一个书店老板，他的书店开了&nbsp;<code>n</code>&nbsp;分钟。每分钟都有一些顾客进入这家商店。给定一个长度为 <code>n</code> 的整数数组 <code>customers</code> ，其中 <code>customers[i]</code> 是在第 <code>i</code> 分钟开始时进入商店的顾客的编号，所有这些顾客在第 <code>i</code> 分钟结束后离开。</p>\n\n<p>在某些时候，书店老板会生气。 如果书店老板在第 <code>i</code> 分钟生气，那么 <code>grumpy[i] = 1</code>，否则 <code>grumpy[i] = 0</code>。</p>\n\n<p>当书店老板生气时，那一分钟的顾客就会不满意，若老板不生气则顾客是满意的。</p>\n\n<p>书店老板知道一个秘密技巧，能抑制自己的情绪，可以让自己连续&nbsp;<code>minutes</code>&nbsp;分钟不生气，但却只能使用一次。</p>\n\n<p>请你返回 <em>这一天营业下来，最多有多少客户能够感到满意</em> 。<br />\n&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], minutes = 3\n<strong>输出：</strong>16\n<strong>解释：</strong>书店老板在最后 3 分钟保持冷静。\n感到满意的最大客户数量 = 1 + 1 + 1 + 1 + 7 + 5 = 16.\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>customers = [1], grumpy = [0], minutes = 1\n<strong>输出：</strong>1</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == customers.length == grumpy.length</code></li>\n\t<li><code>1 &lt;= minutes &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= customers[i] &lt;= 1000</code></li>\n\t<li><code>grumpy[i] == 0 or 1</code></li>\n</ul>\n",
             "isPaidOnly": false,
             "difficulty": "Medium",
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@@ -143,7 +143,7 @@
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             "hints": [
                 "Say the store owner uses their power in minute 1 to X and we have some answer A.  If they instead use their power from minute 2 to X+1, we only have to use data from minutes 1, 2, X and X+1 to update our answer A."
             ],