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].
题目大意：给2n个数，请把数字分为2个一组，问所有组（取每组数的较小的那一数字）累加的和最大为多少～
分析：把数组从小到大排列，第1个、第3个、…第2n-1个数字之和即为所求～
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class Solution { public: int arrayPairSum(vector<int>& nums) { sort(nums.begin(), nums.end()); int ans = 0; for (int i = 0; i < nums.size(); i+=2) ans += nums[i]; return ans; } }; |
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