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leetcode-cn/originData/max-consecutive-ones-iii.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给定一个二进制数组&nbsp;<code>nums</code>&nbsp;和一个整数 <code>k</code>，如果可以翻转最多 <code>k</code> 个 <code>0</code> ，则返回 <em>数组中连续 <code>1</code> 的最大个数</em> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,1,1,0,0,0,1,1,1,1,0], K = 2\n<strong>输出：</strong>6\n<strong>解释：</strong>[1,1,1,0,0,<strong>1</strong>,1,1,1,1,<strong>1</strong>]\n粗体数字从 0 翻转到 1，最长的子数组长度为 6。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3\n<strong>输出：</strong>10\n<strong>解释：</strong>[0,0,1,1,<strong>1</strong>,<strong>1</strong>,1,1,1,<strong>1</strong>,1,1,0,0,0,1,1,1,1]\n粗体数字从 0 翻转到 1，最长的子数组长度为 10。</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;= 10<sup>5</sup></code></li>\n\t<li><code>nums[i]</code>&nbsp;不是&nbsp;<code>0</code>&nbsp;就是&nbsp;<code>1</code></li>\n\t<li><code>0 &lt;= k &lt;= nums.length</code></li>\n</ul>\n",
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             "hints": [
                 "One thing's for sure, we will only flip a zero if it extends an existing window of 1s. Otherwise, there's no point in doing it, right? Think Sliding Window!",
                 "Since we know this problem can be solved using the sliding window construct, we might as well focus in that direction for hints. Basically, in a given window, we can never have > K zeros, right?",