1877. Minimize Maximum Pair Sum in Array【贪心/排序】

The pair sum of a pair (a,b) is equal to a + b. The maximum pair sum is the largest pair sum in a list of pairs.

• For example, if we have pairs (1,5), (2,3), and (4,4), the maximum pair sum would be max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8.

Given an array nums of even length n, pair up the elements of nums into n / 2 pairs such that:

• Each element of nums is in exactly one pair, and
• The maximum pair sum is minimized.

Return the minimized maximum pair sum after optimally pairing up the elements.

Example 1:

Input: nums = [3,5,2,3]
Output: 7
Explanation: The elements can be paired up into pairs (3,3) and (5,2).
The maximum pair sum is max(3+3, 5+2) = max(6, 7) = 7. 

Example 2:

Input: nums = [3,5,4,2,4,6]
Output: 8
Explanation: The elements can be paired up into pairs (3,5), (4,4), and (6,2).
The maximum pair sum is max(3+5, 4+4, 6+2) = max(8, 8, 8) = 8.

Constraints:

• n == nums.length
• 2 <= n <= 105
• n is even.
• 1 <= nums[i] <= 105

题意：一个数对 (a,b) 的 数对和 等于 a + b 。最大数对和 是一个数对数组中最大的 数对和 。

给你一个长度为 偶数 n 的数组 nums ，请你将 nums 中的元素分成 n / 2 个数对，在最优数对划分的方案下，返回最小的 最大数对和 。

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解法 贪心+排序

从小到大排序，最大的元素和最小的元素匹配，依次配对形成数对，返回最大的数对和即可。

class Solution {
public:
    int minPairSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int ans = 0;
        for (int lo = 0, hi = nums.size() - 1; lo < hi; ++lo, --hi) ans = max(ans, nums[lo] + nums[hi]);
        return ans;
    }
};

运行效率如下：

执行用时：244 ms, 在所有 C++ 提交中击败了92.02% 的用户
内存消耗：94 MB, 在所有 C++ 提交中击败了75.19% 的用户