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leetcode-cn/originData/burst-balloons.json
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@@ -9,10 +9,10 @@
"titleSlug": "burst-balloons",
"content": "<p>You are given <code>n</code> balloons, indexed from <code>0</code> to <code>n - 1</code>. Each balloon is painted with a number on it represented by an array <code>nums</code>. You are asked to burst all the balloons.</p>\n\n<p>If you burst the <code>i<sup>th</sup></code> balloon, you will get <code>nums[i - 1] * nums[i] * nums[i + 1]</code> coins. If <code>i - 1</code> or <code>i + 1</code> goes out of bounds of the array, then treat it as if there is a balloon with a <code>1</code> painted on it.</p>\n\n<p>Return <em>the maximum coins you can collect by bursting the balloons wisely</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,1,5,8]\n<strong>Output:</strong> 167\n<strong>Explanation:</strong>\nnums = [3,1,5,8] --&gt; [3,5,8] --&gt; [3,8] --&gt; [8] --&gt; []\ncoins = 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,5]\n<strong>Output:</strong> 10\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 300</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 100</code></li>\n</ul>\n",
"translatedTitle": "戳气球",
"translatedContent": "<p>有 <code>n</code> 个气球，编号为<code>0</code> 到 <code>n - 1</code>，每个气球上都标有一个数字，这些数字存在数组 <code>nums</code> 中。</p>\n\n<p>现在要求你戳破所有的气球。戳破第 <code>i</code> 个气球，你可以获得 <code>nums[i - 1] * nums[i] * nums[i + 1]</code> 枚硬币。 这里的 <code>i - 1</code> 和 <code>i + 1</code> 代表和 <code>i</code> 相邻的两个气球的序号。如果 <code>i - 1</code>或 <code>i + 1</code> 超出了数组的边界，那么就当它是一个数字为 <code>1</code> 的气球。</p>\n\n<p>求所能获得硬币的最大数量。</p>\n\n<p> </p>\n<strong>示例 1：</strong>\n\n<pre>\n<strong>输入：</strong>nums = [3,1,5,8]\n<strong>输出：</strong>167\n<strong>解释：</strong>\nnums = [3,1,5,8] --> [3,5,8] --> [3,8] --> [8] --> []\ncoins = 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,5]\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>n == nums.length</code></li>\n\t<li><code>1 <= n <= 500</code></li>\n\t<li><code>0 <= nums[i] <= 100</code></li>\n</ul>\n",
"translatedContent": "<p>有 <code>n</code> 个气球，编号为<code>0</code> 到 <code>n - 1</code>，每个气球上都标有一个数字，这些数字存在数组&nbsp;<code>nums</code>&nbsp;中。</p>\n\n<p>现在要求你戳破所有的气球。戳破第 <code>i</code> 个气球，你可以获得&nbsp;<code>nums[i - 1] * nums[i] * nums[i + 1]</code> 枚硬币。&nbsp;这里的 <code>i - 1</code> 和 <code>i + 1</code> 代表和&nbsp;<code>i</code>&nbsp;相邻的两个气球的序号。如果 <code>i - 1</code>或 <code>i + 1</code> 超出了数组的边界，那么就当它是一个数字为 <code>1</code> 的气球。</p>\n\n<p>求所能获得硬币的最大数量。</p>\n\n<p>&nbsp;</p>\n<strong>示例 1：</strong>\n\n<pre>\n<strong>输入：</strong>nums = [3,1,5,8]\n<strong>输出：</strong>167\n<strong>解释：</strong>\nnums = [3,1,5,8] --&gt; [3,5,8] --&gt; [3,8] --&gt; [8] --&gt; []\ncoins = 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,5]\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>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 300</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 100</code></li>\n</ul>\n",
"isPaidOnly": false,
"difficulty": "Hard",
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"similarQuestions": "[{\"title\": \"Minimum Cost to Merge Stones\", \"titleSlug\": \"minimum-cost-to-merge-stones\", \"difficulty\": \"Hard\", \"translatedTitle\": \"\\u5408\\u5e76\\u77f3\\u5934\\u7684\\u6700\\u4f4e\\u6210\\u672c\"}]",
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