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leetcode-cn/originData/minimum-size-subarray-sum.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给定一个含有 <code>n</code><strong> </strong>个正整数的数组和一个正整数 <code>target</code><strong> 。</strong></p>\n\n<p>找出该数组中满足其和<strong> </strong><code>≥ target</code><strong> </strong>的长度最小的 <strong>连续子数组</strong> <code>[nums<sub>l</sub>, nums<sub>l+1</sub>, ..., nums<sub>r-1</sub>, nums<sub>r</sub>]</code> ，并返回其长度<strong>。</strong>如果不存在符合条件的子数组，返回 <code>0</code> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>target = 7, nums = [2,3,1,2,4,3]\n<strong>输出：</strong>2\n<strong>解释：</strong>子数组 <code>[4,3]</code> 是该条件下的长度最小的子数组。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>target = 4, nums = [1,4,4]\n<strong>输出：</strong>1\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>target = 11, nums = [1,1,1,1,1,1,1,1]\n<strong>输出：</strong>0\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= target <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>5</sup></code></li>\n</ul>\n\n<p> </p>\n\n<p><strong>进阶：</strong></p>\n\n<ul>\n\t<li>如果你已经实现<em> </em><code>O(n)</code> 时间复杂度的解法, 请尝试设计一个 <code>O(n log(n))</code> 时间复杂度的解法。</li>\n</ul>\n",
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