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leetcode-cn/problem (Chinese)/数组拆分 [array-partition].html

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+<p>给定长度为&nbsp;<code>2n</code><strong>&nbsp;</strong>的整数数组 <code>nums</code> ，你的任务是将这些数分成&nbsp;<code>n</code><strong> </strong>对, 例如 <code>(a<sub>1</sub>, b<sub>1</sub>), (a<sub>2</sub>, b<sub>2</sub>), ..., (a<sub>n</sub>, b<sub>n</sub>)</code> ，使得从 <code>1</code> 到&nbsp;<code>n</code> 的 <code>min(a<sub>i</sub>, b<sub>i</sub>)</code> 总和最大。</p>
+
+<p>返回该 <strong>最大总和</strong> 。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1,4,3,2]
+<strong>输出：</strong>4
+<strong>解释：</strong>所有可能的分法（忽略元素顺序）为：
+1. (1, 4), (2, 3) -&gt; min(1, 4) + min(2, 3) = 1 + 2 = 3
+2. (1, 3), (2, 4) -&gt; min(1, 3) + min(2, 4) = 1 + 2 = 3
+3. (1, 2), (3, 4) -&gt; min(1, 2) + min(3, 4) = 1 + 3 = 4
+所以最大总和为 4</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [6,2,6,5,1,2]
+<strong>输出：</strong>9
+<strong>解释：</strong>最优的分法为 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
+	<li><code>nums.length == 2 * n</code></li>
+	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
+</ul>