leetcode-problemset/leetcode-cn/originData/maximum-subarray.json

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            "title": "Maximum Subarray",
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            "content": "<p>Given an integer array <code>nums</code>, find the <span data-keyword=\"subarray-nonempty\">subarray</span> with the largest sum, and return <em>its sum</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [-2,1,-3,4,-1,2,1,-5,4]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> The subarray [4,-1,2,1] has the largest sum 6.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> The subarray [1] has the largest sum 1.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [5,4,-1,7,8]\n<strong>Output:</strong> 23\n<strong>Explanation:</strong> The subarray [5,4,-1,7,8] has the largest sum 23.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up:</strong> If you have figured out the <code>O(n)</code> solution, try coding another solution using the <strong>divide and conquer</strong> approach, which is more subtle.</p>\n",
            "translatedTitle": "最大子数组和",
            "translatedContent": "<p>给你一个整数数组 <code>nums</code> ，请你找出一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。</p>\n\n<p><strong>子数组 </strong>是数组中的一个连续部分。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [-2,1,-3,4,-1,2,1,-5,4]\n<strong>输出：</strong>6\n<strong>解释：</strong>连续子数组&nbsp;[4,-1,2,1] 的和最大，为&nbsp;6 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1]\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>nums = [5,4,-1,7,8]\n<strong>输出：</strong>23\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;= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>如果你已经实现复杂度为 <code>O(n)</code> 的解法，尝试使用更为精妙的 <strong>分治法</strong> 求解。</p>\n",
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