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leetcode-cn/originData/maximum-sum-of-3-non-overlapping-subarrays.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code> ，找出三个长度为 <code>k</code> 、互不重叠、且全部数字和（<code>3 * k</code> 项）最大的子数组，并返回这三个子数组。</p>\n\n<p>以下标的数组形式返回结果，数组中的每一项分别指示每个子数组的起始位置（下标从 <strong>0</strong> 开始）。如果有多个结果，返回字典序最小的一个。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,1,2,6,7,5,1], k = 2\n<strong>输出：</strong>[0,3,5]\n<strong>解释：</strong>子数组 [1, 2], [2, 6], [7, 5] 对应的起始下标为 [0, 3, 5]。\n也可以取 [2, 1], 但是结果 [1, 3, 5] 在字典序上更大。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,1,2,1,2,1,2,1], k = 2\n<strong>输出：</strong>[0,2,4]\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;= 2 * 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;&nbsp;2<sup>16</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= floor(nums.length / 3)</code></li>\n</ul>\n",
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