leetcode-problemset/leetcode-cn/originData/find-the-winner-of-the-circular-game.json

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            "title": "Find the Winner of the Circular Game",
            "titleSlug": "find-the-winner-of-the-circular-game",
            "content": "<p>There are <code>n</code> friends that are playing a game. The friends are sitting in a circle and are numbered from <code>1</code> to <code>n</code> in <strong>clockwise order</strong>. More formally, moving clockwise from the <code>i<sup>th</sup></code> friend brings you to the <code>(i+1)<sup>th</sup></code> friend for <code>1 &lt;= i &lt; n</code>, and moving clockwise from the <code>n<sup>th</sup></code> friend brings you to the <code>1<sup>st</sup></code> friend.</p>\n\n<p>The rules of the game are as follows:</p>\n\n<ol>\n\t<li><strong>Start</strong> at the <code>1<sup>st</sup></code> friend.</li>\n\t<li>Count the next <code>k</code> friends in the clockwise direction <strong>including</strong> the friend you started at. The counting wraps around the circle and may count some friends more than once.</li>\n\t<li>The last friend you counted leaves the circle and loses the game.</li>\n\t<li>If there is still more than one friend in the circle, go back to step <code>2</code> <strong>starting</strong> from the friend <strong>immediately clockwise</strong> of the friend who just lost and repeat.</li>\n\t<li>Else, the last friend in the circle wins the game.</li>\n</ol>\n\n<p>Given the number of friends, <code>n</code>, and an integer <code>k</code>, return <em>the winner of the game</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/03/25/ic234-q2-ex11.png\" style=\"width: 500px; height: 345px;\" />\n<pre>\n<strong>Input:</strong> n = 5, k = 2\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> Here are the steps of the game:\n1) Start at friend 1.\n2) Count 2 friends clockwise, which are friends 1 and 2.\n3) Friend 2 leaves the circle. Next start is friend 3.\n4) Count 2 friends clockwise, which are friends 3 and 4.\n5) Friend 4 leaves the circle. Next start is friend 5.\n6) Count 2 friends clockwise, which are friends 5 and 1.\n7) Friend 1 leaves the circle. Next start is friend 3.\n8) Count 2 friends clockwise, which are friends 3 and 5.\n9) Friend 5 leaves the circle. Only friend 3 is left, so they are the winner.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 6, k = 5\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> The friends leave in this order: 5, 4, 6, 2, 3. The winner is friend 1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 500</code></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up:</strong></p>\n\n<p>Could you solve this problem in linear time with constant space?</p>\n",
            "translatedTitle": "找出游戏的获胜者",
            "translatedContent": "<p>共有 <code>n</code> 名小伙伴一起做游戏。小伙伴们围成一圈，按 <strong>顺时针顺序</strong> 从 <code>1</code> 到 <code>n</code> 编号。确切地说，从第 <code>i</code> 名小伙伴顺时针移动一位会到达第 <code>(i+1)</code> 名小伙伴的位置，其中 <code>1 &lt;= i &lt; n</code> ，从第 <code>n</code> 名小伙伴顺时针移动一位会回到第 <code>1</code> 名小伙伴的位置。</p>\n\n<p>游戏遵循如下规则：</p>\n\n<ol>\n\t<li>从第 <code>1</code> 名小伙伴所在位置 <strong>开始</strong> 。</li>\n\t<li>沿着顺时针方向数 <code>k</code> 名小伙伴，计数时需要 <strong>包含</strong> 起始时的那位小伙伴。逐个绕圈进行计数，一些小伙伴可能会被数过不止一次。</li>\n\t<li>你数到的最后一名小伙伴需要离开圈子，并视作输掉游戏。</li>\n\t<li>如果圈子中仍然有不止一名小伙伴，从刚刚输掉的小伙伴的 <strong>顺时针下一位</strong> 小伙伴 <strong>开始</strong>，回到步骤 <code>2</code> 继续执行。</li>\n\t<li>否则，圈子中最后一名小伙伴赢得游戏。</li>\n</ol>\n\n<p>给你参与游戏的小伙伴总数 <code>n</code> ，和一个整数 <code>k</code> ，返回游戏的获胜者。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/03/25/ic234-q2-ex11.png\" style=\"width: 500px; height: 345px;\" />\n<pre>\n<strong>输入：</strong>n = 5, k = 2\n<strong>输出：</strong>3\n<strong>解释：</strong>游戏运行步骤如下：\n1) 从小伙伴 1 开始。\n2) 顺时针数 2 名小伙伴，也就是小伙伴 1 和 2 。\n3) 小伙伴 2 离开圈子。下一次从小伙伴 3 开始。\n4) 顺时针数 2 名小伙伴，也就是小伙伴 3 和 4 。\n5) 小伙伴 4 离开圈子。下一次从小伙伴 5 开始。\n6) 顺时针数 2 名小伙伴，也就是小伙伴 5 和 1 。\n7) 小伙伴 1 离开圈子。下一次从小伙伴 3 开始。\n8) 顺时针数 2 名小伙伴，也就是小伙伴 3 和 5 。\n9) 小伙伴 5 离开圈子。只剩下小伙伴 3 。所以小伙伴 3 是游戏的获胜者。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 6, k = 5\n<strong>输出：</strong>1\n<strong>解释：</strong>小伙伴离开圈子的顺序：5、4、6、2、3 。小伙伴 1 是游戏的获胜者。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 500</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你能否使用线性时间复杂度和常数空间复杂度解决此问题？</p>\n",
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