Array Partition I

Posted 2020-10-04 白常福

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:

1. n is a positive integer, which is in the range of [1, 10000].
2. All the integers in the array will be in the range of [-10000, 10000].

 

分析：将给出的整型数组分成两两一组后，返回每组最小值之和，要求和最大。

思路：分组时，让每组的差的平方和最小即可。最简单的方式，直接排序，依次分组。

 JAVA CODE

class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int sum = 0;
        for(int i = 0; i < nums.length; i += 2){
            sum += Math.min(nums[i],nums[i+1]);
        }
        return sum;
    }
}