mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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168 lines
20 KiB
JSON
168 lines
20 KiB
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"title": "Triangle",
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"content": "<p>Given a <code>triangle</code> array, return <em>the minimum path sum from top to bottom</em>.</p>\n\n<p>For each step, you may move to an adjacent number of the row below. More formally, if you are on index <code>i</code> on the current row, you may move to either index <code>i</code> or index <code>i + 1</code> on the next row.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]\n<strong>Output:</strong> 11\n<strong>Explanation:</strong> The triangle looks like:\n <u>2</u>\n <u>3</u> 4\n 6 <u>5</u> 7\n4 <u>1</u> 8 3\nThe minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> triangle = [[-10]]\n<strong>Output:</strong> -10\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= triangle.length <= 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> <= triangle[i][j] <= 10<sup>4</sup></code></li>\n</ul>\n\n<p> </p>\n<strong>Follow up:</strong> Could you do this using only <code>O(n)</code> extra space, where <code>n</code> is the total number of rows in the triangle?",
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"translatedTitle": "三角形最小路径和",
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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> </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 11（即，2 + 3 + 5 + 1 = 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> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= triangle.length <= 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> <= triangle[i][j] <= 10<sup>4</sup></code></li>\n</ul>\n\n<p> </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> 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",
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