mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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195 lines
24 KiB
JSON
195 lines
24 KiB
JSON
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"title": "Stone Game VIII",
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"content": "<p>Alice and Bob take turns playing a game, with <strong>Alice starting first</strong>.</p>\r\n\r\n<p>There are <code>n</code> stones arranged in a row. On each player's turn, while the number of stones is <strong>more than one</strong>, they will do the following:</p>\r\n\r\n<ol>\r\n\t<li>Choose an integer <code>x > 1</code>, and <strong>remove</strong> the leftmost <code>x</code> stones from the row.</li>\r\n\t<li>Add the <strong>sum</strong> of the <strong>removed</strong> stones' values to the player's score.</li>\r\n\t<li>Place a <strong>new stone</strong>, whose value is equal to that sum, on the left side of the row.</li>\r\n</ol>\r\n\r\n<p>The game stops when <strong>only</strong> <strong>one</strong> stone is left in the row.</p>\r\n\r\n<p>The <strong>score difference</strong> between Alice and Bob is <code>(Alice's score - Bob's score)</code>. Alice's goal is to <strong>maximize</strong> the score difference, and Bob's goal is the <strong>minimize</strong> the score difference.</p>\r\n\r\n<p>Given an integer array <code>stones</code> of length <code>n</code> where <code>stones[i]</code> represents the value of the <code>i<sup>th</sup></code> stone <strong>from the left</strong>, return <em>the <strong>score difference</strong> between Alice and Bob if they both play <strong>optimally</strong>.</em></p>\r\n\r\n<p> </p>\r\n<p><strong class=\"example\">Example 1:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> stones = [-1,2,-3,4,-5]\r\n<strong>Output:</strong> 5\r\n<strong>Explanation:</strong>\r\n- Alice removes the first 4 stones, adds (-1) + 2 + (-3) + 4 = 2 to her score, and places a stone of\r\n value 2 on the left. stones = [2,-5].\r\n- Bob removes the first 2 stones, adds 2 + (-5) = -3 to his score, and places a stone of value -3 on\r\n the left. stones = [-3].\r\nThe difference between their scores is 2 - (-3) = 5.\r\n</pre>\r\n\r\n<p><strong class=\"example\">Example 2:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> stones = [7,-6,5,10,5,-2,-6]\r\n<strong>Output:</strong> 13\r\n<strong>Explanation:</strong>\r\n- Alice removes all stones, adds 7 + (-6) + 5 + 10 + 5 + (-2) + (-6) = 13 to her score, and places a\r\n stone of value 13 on the left. stones = [13].\r\nThe difference between their scores is 13 - 0 = 13.\r\n</pre>\r\n\r\n<p><strong class=\"example\">Example 3:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> stones = [-10,-12]\r\n<strong>Output:</strong> -22\r\n<strong>Explanation:</strong>\r\n- Alice can only make one move, which is to remove both stones. She adds (-10) + (-12) = -22 to her\r\n score and places a stone of value -22 on the left. stones = [-22].\r\nThe difference between their scores is (-22) - 0 = -22.\r\n</pre>\r\n\r\n<p> </p>\r\n<p><strong>Constraints:</strong></p>\r\n\r\n<ul>\r\n\t<li><code>n == stones.length</code></li>\r\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\r\n\t<li><code>-10<sup>4</sup> <= stones[i] <= 10<sup>4</sup></code></li>\r\n</ul>",
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"translatedTitle": "石子游戏 VIII",
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"translatedContent": "<p>Alice 和 Bob 玩一个游戏,两人轮流操作, <strong>Alice 先手</strong> 。</p>\n\n<p>总共有 <code>n</code> 个石子排成一行。轮到某个玩家的回合时,如果石子的数目 <strong>大于 1</strong> ,他将执行以下操作:</p>\n\n<ol>\n\t<li>选择一个整数 <code>x > 1</code> ,并且 <strong>移除</strong> 最左边的 <code>x</code> 个石子。</li>\n\t<li>将<strong> 移除</strong> 的石子价值之 <strong>和</strong> 累加到该玩家的分数中。</li>\n\t<li>将一个 <strong>新的石子</strong> 放在最左边,且新石子的值为被移除石子值之和。</li>\n</ol>\n\n<p>当只剩下 <strong>一个</strong> 石子时,游戏结束。</p>\n\n<p>Alice 和 Bob 的 <strong>分数之差</strong> 为 <code>(Alice 的分数 - Bob 的分数)</code> 。 Alice 的目标是<strong> 最大化</strong> 分数差,Bob 的目标是 <strong>最小化</strong> 分数差。</p>\n\n<p>给你一个长度为 <code>n</code> 的整数数组 <code>stones</code> ,其中 <code>stones[i]</code> 是 <strong>从左边起</strong> 第 <code>i</code> 个石子的价值。请你返回在双方都采用 <strong>最优</strong> 策略的情况下,Alice 和 Bob 的 <strong>分数之差</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre><b>输入:</b>stones = [-1,2,-3,4,-5]\n<b>输出:</b>5\n<strong>解释:</strong>\n- Alice 移除最左边的 4 个石子,得分增加 (-1) + 2 + (-3) + 4 = 2 ,并且将一个价值为 2 的石子放在最左边。stones = [2,-5] 。\n- Bob 移除最左边的 2 个石子,得分增加 2 + (-5) = -3 ,并且将一个价值为 -3 的石子放在最左边。stones = [-3] 。\n两者分数之差为 2 - (-3) = 5 。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre><b>输入:</b>stones = [7,-6,5,10,5,-2,-6]\n<b>输出:</b>13\n<b>解释:</b>\n- Alice 移除所有石子,得分增加 7 + (-6) + 5 + 10 + 5 + (-2) + (-6) = 13 ,并且将一个价值为 13 的石子放在最左边。stones = [13] 。\n两者分数之差为 13 - 0 = 13 。\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre><b>输入:</b>stones = [-10,-12]\n<b>输出:</b>-22\n<strong>解释:</strong>\n- Alice 只有一种操作,就是移除所有石子。得分增加 (-10) + (-12) = -22 ,并且将一个价值为 -22 的石子放在最左边。stones = [-22] 。\n两者分数之差为 (-22) - 0 = -22 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>n == stones.length</code></li>\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>4</sup> <= stones[i] <= 10<sup>4</sup></code></li>\n</ul>\n",
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"Let's note that the only thing that matters is how many stones were removed so we can maintain dp[numberOfRemovedStones]",
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