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package leetcode.easy;
import java.util.Arrays;
/**
* @author chen_wj
* @Description:
* @date 2021/2/16
* @Description:
* 给定长度为 2n 的整数数组 nums ,你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ,使得从 1 到 n 的 min(ai, bi) 总和最大。
*
* 返回该 最大总和 。
*
*
*
* 示例 1:
*
* 输入:nums = [1,4,3,2]
* 输出:4
* 解释:所有可能的分法(忽略元素顺序)为:
* 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
* 所以最大总和为 4
* 示例 2:
*
* 输入:nums = [6,2,6,5,1,2]
* 输出:9
* 解释:最优的分法为 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9
*
*
* 提示:
*
* 1 <= n <= 104
* nums.length == 2 * n
* -104 <= nums[i] <= 104
*
* 来源:力扣(LeetCode)
* 链接:https://leetcode-cn.com/problems/array-partition-i
* 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
* @modifier
*/
public class Array_Partition {
public static void main(String[] args) {
int[] arr = {6,2,6,5,1,2};
System.out.println(arrayPairSum(arr));
}
public static int arrayPairSum(int[] nums) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i<nums.length; i += 2) {
sum+=nums[i];
}
return sum;
}
}