Array Partition
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. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
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 <= 104
nums.length == 2 * n
-104 <= nums[i] <= 104
思路一：观察数组排列，由于要从数组中取一半的数字求和，这一半的数字每一个都要小于另外一个数字，所以最好的情况是排序后，取偶数下标的数字。例如下面，最好的情况是取 1+3+5
1--2--3--4--5--6
public int arrayPairSum(int[] nums) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length; i+=2) {
sum += nums[i];
}
return sum;
}