mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
183 lines
22 KiB
JSON
183 lines
22 KiB
JSON
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"boundTopicId": 86519,
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"title": "Jump Game IV",
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"content": "<p>Given an array of integers <code>arr</code>, you are initially positioned at the first index of the array.</p>\n\n<p>In one step you can jump from index <code>i</code> to index:</p>\n\n<ul>\n\t<li><code>i + 1</code> where: <code>i + 1 < arr.length</code>.</li>\n\t<li><code>i - 1</code> where: <code>i - 1 >= 0</code>.</li>\n\t<li><code>j</code> where: <code>arr[i] == arr[j]</code> and <code>i != j</code>.</li>\n</ul>\n\n<p>Return <em>the minimum number of steps</em> to reach the <strong>last index</strong> of the array.</p>\n\n<p>Notice that you can not jump outside of the array at any time.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [100,-23,-23,404,100,23,23,23,3,404]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [7]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Start index is the last index. You do not need to jump.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [7,6,9,6,9,6,9,7]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> You can jump directly from index 0 to index 7 which is last index of the array.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= arr.length <= 5 * 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>8</sup> <= arr[i] <= 10<sup>8</sup></code></li>\n</ul>\n",
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"translatedTitle": "跳跃游戏 IV",
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"translatedContent": "<p>给你一个整数数组 <code>arr</code> ，你一开始在数组的第一个元素处（下标为 0）。</p>\n\n<p>每一步，你可以从下标 <code>i</code> 跳到下标 <code>i + 1</code> 、<code>i - 1</code> 或者 <code>j</code> ：</p>\n\n<ul>\n\t<li><code>i + 1</code> 需满足：<code>i + 1 < arr.length</code></li>\n\t<li><code>i - 1</code> 需满足：<code>i - 1 >= 0</code></li>\n\t<li><code>j</code> 需满足：<code>arr[i] == arr[j]</code> 且 <code>i != j</code></li>\n</ul>\n\n<p>请你返回到达数组最后一个元素的下标处所需的 <strong>最少操作次数</strong> 。</p>\n\n<p>注意：任何时候你都不能跳到数组外面。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>arr = [100,-23,-23,404,100,23,23,23,3,404]\n<strong>输出：</strong>3\n<strong>解释：</strong>那你需要跳跃 3 次，下标依次为 0 --> 4 --> 3 --> 9 。下标 9 为数组的最后一个元素的下标。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>arr = [7]\n<strong>输出：</strong>0\n<strong>解释：</strong>一开始就在最后一个元素处，所以你不需要跳跃。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>arr = [7,6,9,6,9,6,9,7]\n<strong>输出：</strong>1\n<strong>解释：</strong>你可以直接从下标 0 处跳到下标 7 处，也就是数组的最后一个元素处。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n<meta charset=\"UTF-8\" />\n\n<ul>\n\t<li><code>1 <= arr.length <= 5 * 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>8</sup> <= arr[i] <= 10<sup>8</sup></code></li>\n</ul>\n",
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"Build a graph of n nodes where nodes are the indices of the array and edges for node i are nodes i+1, i-1, j where arr[i] == arr[j].",
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"Start bfs from node 0 and keep distance. The answer is the distance when you reach node n-1."
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