LeetCode | 561. Array Partition I

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题目

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.

Example 1:

Input: nums = [1,4,3,2]
Output: 4
Explanation: All possible pairings (ignoring the ordering of elements) are:

(1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
(1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
(1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
So the maximum possible sum is 4.

Example 2:

Input: nums = [6,2,6,5,1,2]
Output: 9
Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

Constraints:

1 <= n <= 10^4
nums.length == 2 * n
-10^4 <= nums[i] <= 10^4

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代码

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        vector<int>::size_type n = nums.size();
		if(n == 0)
			return 0;
		sort(nums.begin(), nums.end());
		int sum = 0;
		for(int i = 0; i<n; i+=2){
			sum += nums[i];
		}
		return sum;
    }
};