mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
177 lines
20 KiB
JSON
177 lines
20 KiB
JSON
{
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"questionId": "2001",
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"questionFrontendId": "1871",
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"categoryTitle": "Algorithms",
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"boundTopicId": 785046,
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"title": "Jump Game VII",
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"titleSlug": "jump-game-vii",
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"content": "<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> and two integers <code>minJump</code> and <code>maxJump</code>. In the beginning, you are standing at index <code>0</code>, which is equal to <code>'0'</code>. You can move from index <code>i</code> to index <code>j</code> if the following conditions are fulfilled:</p>\n\n<ul>\n\t<li><code>i + minJump <= j <= min(i + maxJump, s.length - 1)</code>, and</li>\n\t<li><code>s[j] == '0'</code>.</li>\n</ul>\n\n<p>Return <code>true</code><i> if you can reach index </i><code>s.length - 1</code><i> in </i><code>s</code><em>, or </em><code>false</code><em> otherwise.</em></p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "<u>0</u>11<u>0</u>1<u>0</u>", minJump = 2, maxJump = 3\n<strong>Output:</strong> true\n<strong>Explanation:</strong>\nIn the first step, move from index 0 to index 3. \nIn the second step, move from index 3 to index 5.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "01101110", minJump = 2, maxJump = 3\n<strong>Output:</strong> false\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= s.length <= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> is either <code>'0'</code> or <code>'1'</code>.</li>\n\t<li><code>s[0] == '0'</code></li>\n\t<li><code>1 <= minJump <= maxJump < s.length</code></li>\n</ul>\n",
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"translatedTitle": "跳跃游戏 VII",
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"translatedContent": "<p>给你一个下标从 <strong>0 </strong>开始的二进制字符串 <code>s</code> 和两个整数 <code>minJump</code> 和 <code>maxJump</code> 。一开始，你在下标 <code>0</code> 处，且该位置的值一定为 <code>'0'</code> 。当同时满足如下条件时，你可以从下标 <code>i</code> 移动到下标 <code>j</code> 处：</p>\n\n<ul>\n\t<li><code>i + minJump <= j <= min(i + maxJump, s.length - 1)</code> 且</li>\n\t<li><code>s[j] == '0'</code>.</li>\n</ul>\n\n<p>如果你可以到达 <code>s</code> 的下标<i> </i><code>s.length - 1</code> 处，请你返回 <code>true</code> ，否则返回 <code>false</code> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"<strong>0</strong>11<strong>0</strong>1<strong>0</strong>\", minJump = 2, maxJump = 3\n<b>输出：</b>true\n<strong>解释：</strong>\n第一步，从下标 0 移动到下标 3 。\n第二步，从下标 3 移动到下标 5 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"01101110\", minJump = 2, maxJump = 3\n<b>输出：</b>false\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 <= s.length <= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> 要么是 <code>'0'</code> ，要么是 <code>'1'</code></li>\n\t<li><code>s[0] == '0'</code></li>\n\t<li><code>1 <= minJump <= maxJump < s.length</code></li>\n</ul>\n",
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"code": "# @param {String} s\n# @param {Integer} min_jump\n# @param {Integer} max_jump\n# @return {Boolean}\ndef can_reach(s, min_jump, max_jump)\n\nend",
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"Consider for each reachable index i the interval [i + a, i + b].",
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"Use partial sums to mark the intervals as reachable."
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