国外版

算法题（国外版）/last-stone-weight-ii.html

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+<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>
+
+<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>
+
+<ul>
+	<li>If <code>x == y</code>, both stones are destroyed, and</li>
+	<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>
+</ul>
+
+<p>At the end of the game, there is <strong>at most one</strong> stone left.</p>
+
+<p>Return <em>the smallest possible weight of the left stone</em>. If there are no stones left, return <code>0</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> stones = [2,7,4,1,8,1]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong>
+We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
+we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
+we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
+we can combine 1 and 1 to get 0, so the array converts to [1], then that&#39;s the optimal value.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> stones = [31,26,33,21,40]
+<strong>Output:</strong> 5
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= stones.length &lt;= 30</code></li>
+	<li><code>1 &lt;= stones[i] &lt;= 100</code></li>
+</ul>