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[LeetCode] Array Partition I
Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
Example 1:
Input: [1,4,3,2] Output: 4 Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
Note:
- n is a positive integer, which is in the range of [1, 10000].
- All the integers in the array will be in the range of [-10000, 10000].
给定一个含有2n个整数的数组,将这2n个整数分成n个整数对,对每个整数对中的较小数求和,并使这个和最大。根据题意,要使总和最大,在整个数组中需要每个整数对中较小数的取值尽可能大。先将数组排序,此时把两两相邻的整数作为一个整数对,求每个整数对较小数的和,即求排序后数组奇数次序的数值之和。
class Solution { public: int arrayPairSum(vector<int>& nums) { sort(nums.begin(), nums.end()); int sum = 0; for (int i = 0; i != nums.size(); i += 2) sum += nums[i]; return sum; } }; // 98 ms
[LeetCode] Array Partition I
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