mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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190 lines
25 KiB
JSON
190 lines
25 KiB
JSON
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"questionId": "2477",
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"questionFrontendId": "2400",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1793491,
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"title": "Number of Ways to Reach a Position After Exactly k Steps",
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"content": "<p>You are given two <strong>positive</strong> integers <code>startPos</code> and <code>endPos</code>. Initially, you are standing at position <code>startPos</code> on an <strong>infinite</strong> number line. With one step, you can move either one position to the left, or one position to the right.</p>\n\n<p>Given a positive integer <code>k</code>, return <em>the number of <strong>different</strong> ways to reach the position </em><code>endPos</code><em> starting from </em><code>startPos</code><em>, such that you perform <strong>exactly</strong> </em><code>k</code><em> steps</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>Two ways are considered different if the order of the steps made is not exactly the same.</p>\n\n<p><strong>Note</strong> that the number line includes negative integers.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> startPos = 1, endPos = 2, k = 3\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> We can reach position 2 from 1 in exactly 3 steps in three ways:\n- 1 -> 2 -> 3 -> 2.\n- 1 -> 2 -> 1 -> 2.\n- 1 -> 0 -> 1 -> 2.\nIt can be proven that no other way is possible, so we return 3.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> startPos = 2, endPos = 5, k = 10\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> It is impossible to reach position 5 from position 2 in exactly 10 steps.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= startPos, endPos, k <= 1000</code></li>\n</ul>\n",
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"translatedTitle": "恰好移动 k 步到达某一位置的方法数目",
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"translatedContent": "<p>给你两个 <strong>正</strong> 整数 <code>startPos</code> 和 <code>endPos</code> 。最初,你站在 <strong>无限</strong> 数轴上位置 <code>startPos</code> 处。在一步移动中,你可以向左或者向右移动一个位置。</p>\n\n<p>给你一个正整数 <code>k</code> ,返回从 <code>startPos</code> 出发、<strong>恰好</strong> 移动 <code>k</code> 步并到达 <code>endPos</code> 的 <strong>不同</strong> 方法数目。由于答案可能会很大,返回对 <code>10<sup>9</sup> + 7</code> <strong>取余</strong> 的结果。</p>\n\n<p>如果所执行移动的顺序不完全相同,则认为两种方法不同。</p>\n\n<p><strong>注意:</strong>数轴包含负整数<strong>。</strong></p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre><strong>输入:</strong>startPos = 1, endPos = 2, k = 3\n<strong>输出:</strong>3\n<strong>解释:</strong>存在 3 种从 1 到 2 且恰好移动 3 步的方法:\n- 1 -> 2 -> 3 -> 2.\n- 1 -> 2 -> 1 -> 2.\n- 1 -> 0 -> 1 -> 2.\n可以证明不存在其他方法,所以返回 3 。</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre><strong>输入:</strong>startPos = 2, endPos = 5, k = 10\n<strong>输出:</strong>0\n<strong>解释:</strong>不存在从 2 到 5 且恰好移动 10 步的方法。</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= startPos, endPos, k <= 1000</code></li>\n</ul>\n",
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target=\\\"_blank\\\">\\u300c\\u4ed3\\u9889\\u7f16\\u7a0b\\u8bed\\u8a00\\u5f00\\u53d1\\u6307\\u5357\\u300d<\\/a><\\/p>\"]}",
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