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"translatedContent": "<p>给定一个三角形 <code>triangle</code> ,找出自顶向下的最小路径和。</p>\n\n<p>每一步只能移动到下一行中相邻的结点上。<strong>相邻的结点 </strong>在这里指的是 <strong>下标</strong> 与 <strong>上一层结点下标</strong> 相同或者等于 <strong>上一层结点下标 + 1</strong> 的两个结点。也就是说,如果正位于当前行的下标 <code>i</code> ,那么下一步可以移动到下一行的下标 <code>i</code> 或 <code>i + 1</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]\n<strong>输出:</strong>11\n<strong>解释:</strong>如下面简图所示:\n <strong>2</strong>\n <strong>3</strong> 4\n 6 <strong>5</strong> 7\n4 <strong>1</strong> 8 3\n自顶向下的最小路径和为&nbsp;112&nbsp;+&nbsp;3&nbsp;+&nbsp;5&nbsp;+&nbsp;1&nbsp;= 11。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>triangle = [[-10]]\n<strong>输出:</strong>-10\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= triangle.length &lt;= 200</code></li>\n\t<li><code>triangle[0].length == 1</code></li>\n\t<li><code>triangle[i].length == triangle[i - 1].length + 1</code></li>\n\t<li><code>-10<sup>4</sup> &lt;= triangle[i][j] &lt;= 10<sup>4</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶:</strong></p>\n\n<ul>\n\t<li>你可以只使用 <code>O(n)</code>&nbsp;的额外空间(<code>n</code> 为三角形的总行数)来解决这个问题吗?</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><meta charset=\"UTF-8\" />注意:本题与主站 120&nbsp;题相同:&nbsp;<a href=\"https://leetcode-cn.com/problems/triangle/\">https://leetcode-cn.com/problems/triangle/</a></p>\n",
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