Power button 561. An array of split

Given an integer array nums of length 2n, your task is to divide these numbers into n pairs, eg (a1, b1), (a2, b2), …, (an, bn) such that from 1 to nThe sum of min(ai, bi) is max.

Returns the max sum .

Example 1:

Input: nums = [1,4,3,2]
Output: 4
Explanation: All possibleThe division method (ignoring the order of elements) is:

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 sum is 4

Example 2:

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

Source: LeetCode
Link: https://leetcode.cn/problems/array-partition

In the n pairs of [a, b] arrays split by the nums array, select a smaller number between a and b, and add them to get a maximum sum
We can see from the first example that isYou can sort the nums array first
We divide the nums array into (nums[0],nums[1]) = (1,2);
(nums[2],nums[3]) = (3,4)
Two pairs, and the largest sum of 1+3 is selected;
That is, nums[0]+nums[2];
In the second example, you can also put the array firstsort [1,2,2,5,6,6]
and the largest sum is nums[0]+nums[2]+nums[4] = 1+2+6=9

Code

class Solution {public int arrayPairSum(int[] nums) {Arrays.sort(nums);int sum = 0;//Start from the first one, add every other for(int i=0;i<nums.length;i=i+2){sum = sum +nums[i];}return sum;}}