Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), ..., (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.
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Example 1:
\n\n\nInput: nums = [1,4,3,2]\nOutput: 4\nExplanation: All possible pairings (ignoring the ordering of elements) are:\n1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3\n2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3\n3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4\nSo the maximum possible sum is 4.\n\n
Example 2:
\n\n\nInput: nums = [6,2,6,5,1,2]\nOutput: 9\nExplanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.\n\n\n
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Constraints:
\n\n1 <= n <= 104nums.length == 2 * n-104 <= nums[i] <= 104给定长度为 2n 的整数数组 nums ，你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ，使得从 1 到 n 的 min(ai, bi) 总和最大。
返回该 最大总和 。
\n\n\n\n
示例 1：
\n\n\n输入：nums = [1,4,3,2]\n输出：4\n解释：所有可能的分法（忽略元素顺序）为：\n1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3\n2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3\n3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4\n所以最大总和为 4\n\n
示例 2：
\n\n\n输入：nums = [6,2,6,5,1,2]\n输出：9\n解释：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\n\n\n
\n\n
提示：
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