leetcode-problemset/leetcode-cn/originData/bag-of-tokens.json

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            "title": "Bag of Tokens",
            "titleSlug": "bag-of-tokens",
            "content": "<p>You have an initial <strong>power</strong> of <code>power</code>, an initial <strong>score</strong> of <code>0</code>, and a bag of <code>tokens</code> where <code>tokens[i]</code> is the value of the <code>i<sup>th</sup></code> token (0-indexed).</p>\n\n<p>Your goal is to maximize your total <strong>score</strong> by potentially playing each token in one of two ways:</p>\n\n<ul>\n\t<li>If your current <strong>power</strong> is at least <code>tokens[i]</code>, you may play the <code>i<sup>th</sup></code> token face up, losing <code>tokens[i]</code> <strong>power</strong> and gaining <code>1</code> <strong>score</strong>.</li>\n\t<li>If your current <strong>score</strong> is at least <code>1</code>, you may play the <code>i<sup>th</sup></code> token face down, gaining <code>tokens[i]</code> <strong>power</strong> and losing <code>1</code> <strong>score</strong>.</li>\n</ul>\n\n<p>Each token may be played <strong>at most</strong> once and <strong>in any order</strong>. You do <strong>not</strong> have to play all the tokens.</p>\n\n<p>Return <em>the largest possible <strong>score</strong> you can achieve after playing any number of tokens</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> tokens = [100], power = 50\n<strong>Output:</strong> 0\n<strong>Explanation</strong><strong>:</strong> Playing the only token in the bag is impossible because you either have too little power or too little score.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> tokens = [100,200], power = 150\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> Play the 0<sup>th</sup> token (100) face up, your power becomes 50 and score becomes 1.\nThere is no need to play the 1<sup>st</sup> token since you cannot play it face up to add to your score.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> tokens = [100,200,300,400], power = 200\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> Play the tokens in this order to get a score of 2:\n1. Play the 0<sup>th</sup> token (100) face up, your power becomes 100 and score becomes 1.\n2. Play the 3<sup>rd</sup> token (400) face down, your power becomes 500 and score becomes 0.\n3. Play the 1<sup>st</sup> token (200) face up, your power becomes 300 and score becomes 1.\n4. Play the 2<sup>nd </sup>token (300) face up, your power becomes 0 and score becomes 2.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= tokens.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= tokens[i],&nbsp;power &lt; 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "令牌放置",
            "translatedContent": "<p>你的初始 <strong>能量</strong> 为 <code>power</code>，初始 <strong>分数</strong> 为&nbsp;<code>0</code>，只有一包令牌 <code>tokens</code> 。其中 <code>tokens[i]</code> 是第 <code>i</code> 个令牌的值（下标从 0 开始）。</p>\n\n<p>令牌可能的两种使用方法如下：</p>\n\n<ul>\n\t<li>如果你至少有&nbsp;<code>token[i]</code>&nbsp;点 <strong>能量</strong> ，可以将令牌 <code>i</code> 置为正面朝上，失去&nbsp;<code>token[i]</code>&nbsp;点 <strong>能量</strong> ，并得到&nbsp;<code>1</code>&nbsp;<strong>分</strong> 。</li>\n\t<li>如果我们至少有&nbsp;<code>1</code>&nbsp;<strong>分 </strong>，可以将令牌 <code>i</code> 置为反面朝上，获得&nbsp;<code>token[i]</code> 点 <strong>能量</strong> ，并失去&nbsp;<code>1</code>&nbsp;<strong>分</strong> 。</li>\n</ul>\n\n<p>每个令牌 <strong>最多</strong> 只能使用一次，使用 <strong>顺序不限</strong> ，<strong>不需</strong> 使用所有令牌。</p>\n\n<p>在使用任意数量的令牌后，返回我们可以得到的最大 <strong>分数</strong> 。</p>\n\n<p>&nbsp;</p>\n\n<ol>\n</ol>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>tokens = [100], power = 50\n<strong>输出：</strong>0\n<strong>解释：</strong>无法使用唯一的令牌，因为能量和分数都太少了。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>tokens = [100,200], power = 150\n<strong>输出：</strong>1\n<strong>解释：</strong>令牌 0 正面朝上，能量变为 50，分数变为 1 。\n不必使用令牌 1 ，因为你无法使用它来提高分数。</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>tokens = [100,200,300,400], power = 200\n<strong>输出：</strong>2\n<strong>解释：</strong>按下面顺序使用令牌可以得到 2 分：\n1. 令牌 0 正面朝上，能量变为 100 ，分数变为 1\n2. 令牌 3 正面朝下，能量变为 500 ，分数变为 0\n3. 令牌 1 正面朝上，能量变为 300 ，分数变为 1\n4. 令牌 2 正面朝上，能量变为 0 ，分数变为 2</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= tokens.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= tokens[i],&nbsp;power &lt; 10<sup>4</sup></code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int bagOfTokensScore(vector<int>& tokens, int power) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def bagOfTokensScore(self, tokens, power):\n        \"\"\"\n        :type tokens: List[int]\n        :type power: int\n        :rtype: int\n        \"\"\"",
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                    "code": "/**\n * @param {number[]} tokens\n * @param {number} power\n * @return {number}\n */\nvar bagOfTokensScore = function(tokens, power) {\n\n};",
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