leetcode-problemset/leetcode-cn/originData/last-stone-weight-ii.json

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        "question": {
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            "categoryTitle": "Algorithms",
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            "title": "Last Stone Weight II",
            "titleSlug": "last-stone-weight-ii",
            "content": "<p>You are given an array of integers <code>stones</code> where <code>stones[i]</code> is the weight of the <code>i<sup>th</sup></code> stone.</p>\n\n<p>We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have weights <code>x</code> and <code>y</code> with <code>x &lt;= y</code>. The result of this smash is:</p>\n\n<ul>\n\t<li>If <code>x == y</code>, both stones are destroyed, and</li>\n\t<li>If <code>x != y</code>, the stone of weight <code>x</code> is destroyed, and the stone of weight <code>y</code> has new weight <code>y - x</code>.</li>\n</ul>\n\n<p>At the end of the game, there is <strong>at most one</strong> stone left.</p>\n\n<p>Return <em>the smallest possible weight of the left stone</em>. If there are no stones left, return <code>0</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> stones = [2,7,4,1,8,1]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong>\nWe can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,\nwe can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,\nwe can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,\nwe can combine 1 and 1 to get 0, so the array converts to [1], then that&#39;s the optimal value.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> stones = [31,26,33,21,40]\n<strong>Output:</strong> 5\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= stones.length &lt;= 30</code></li>\n\t<li><code>1 &lt;= stones[i] &lt;= 100</code></li>\n</ul>\n",
            "translatedTitle": "最后一块石头的重量 II",
            "translatedContent": "<p>有一堆石头，用整数数组&nbsp;<code>stones</code> 表示。其中&nbsp;<code>stones[i]</code> 表示第 <code>i</code> 块石头的重量。</p>\n\n<p>每一回合，从中选出<strong>任意两块石头</strong>，然后将它们一起粉碎。假设石头的重量分别为&nbsp;<code>x</code> 和&nbsp;<code>y</code>，且&nbsp;<code>x &lt;= y</code>。那么粉碎的可能结果如下：</p>\n\n<ul>\n\t<li>如果&nbsp;<code>x == y</code>，那么两块石头都会被完全粉碎；</li>\n\t<li>如果&nbsp;<code>x != y</code>，那么重量为&nbsp;<code>x</code>&nbsp;的石头将会完全粉碎，而重量为&nbsp;<code>y</code>&nbsp;的石头新重量为&nbsp;<code>y-x</code>。</li>\n</ul>\n\n<p>最后，<strong>最多只会剩下一块 </strong>石头。返回此石头 <strong>最小的可能重量 </strong>。如果没有石头剩下，就返回 <code>0</code>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>stones = [2,7,4,1,8,1]\n<strong>输出：</strong>1\n<strong>解释：</strong>\n组合 2 和 4，得到 2，所以数组转化为 [2,7,1,8,1]，\n组合 7 和 8，得到 1，所以数组转化为 [2,1,1,1]，\n组合 2 和 1，得到 1，所以数组转化为 [1,1,1]，\n组合 1 和 1，得到 0，所以数组转化为 [1]，这就是最优值。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>stones = [31,26,33,21,40]\n<strong>输出：</strong>5\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= stones.length &lt;= 30</code></li>\n\t<li><code>1 &lt;= stones[i] &lt;= 100</code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int lastStoneWeightII(vector<int>& stones) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def lastStoneWeightII(self, stones):\n        \"\"\"\n        :type stones: List[int]\n        :rtype: int\n        \"\"\"",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[] $stones\n     * @return Integer\n     */\n    function lastStoneWeightII($stones) {\n\n    }\n}",
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            "hints": [
                "Think of the final answer as a sum of weights with + or - sign symbols infront of each weight.  Actually, all sums with 1 of each sign symbol are possible.",
                "Use dynamic programming: for every possible sum with N stones, those sums +x or -x is possible with N+1 stones, where x is the value of the newest stone.  (This overcounts sums that are all positive or all negative, but those don't matter.)"
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