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描述
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.
Note:
n == nums.length
2 <= n <= 10^5
n is even.
1 <= nums[i] <= 10^5
解析
根据题意，给出一个偶数长度的整数列表 nums ，将里面的元素两两组和成一对，每个元素只能属于一个对，使得所有对的和的最大值最小，问这个最小值是多少。细分析一下其实不难，要想使得最大值最小，肯定要让 nums 中最大的元素和最小的元素进行组队求和，次最大的元素和次最小的元素进行组队求和，所以思路很明显就出来了：
- 初始化一个结果变量 result 为 0
- 对 nums 进行升序排序
- 遍历索引 [0, len(nums)//2] ，比较 nums[i]+nums[-1-i] 和 result ，将较大值赋给 result
- 遍历结束得到的 result 即为答案
解答
class Solution(object):
def minPairSum(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result = 0
N = len(nums)
nums.sort()
j = -1
i = 0
while i<N//2:
result = max(result, nums[i]+nums[j])
i+=1
j-=1
return result
运行结果
Runtime: 1132 ms, faster than 39.81% of Python online submissions for Minimize Maximum Pair Sum in Array.
Memory Usage: 25.9 MB, less than 37.96% of Python online submissions for Minimize Maximum Pair Sum in Array.
原题链接：leetcode.com/problems/mi…
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