LeetCode 561. Array Partition I(Easy)

Description:
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].

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Analysis:
$min\{a_{i}, b_{i}\} = \frac{a_{i} + b_{i}}{2} - \frac{b_{i} - a_{i}}{2}$(Assume $a_{i} \leq b_{i}$)
$sum = \frac{a_{1} + b_{1} + a_{2} + b_{2} + \dots + a_{i}+ b_{i}}{2} - \frac{(b_{1} - a_{1}) + (b_{2} - a_{2}) + \dots +(b_{n} - a_{n})}{2}$
To get the maximum of $sum$, we need to make sure $(b_{1} - a_{1}) + (b_{2} - a_{2}) + \dots +(b_{n} - a_{n})$ is minimal. What’s more, if and only if the array is sorted in an increasing order, $(b_{1} - a_{1}) + (b_{2} - a_{2}) + \dots +(b_{n} - a_{n})$ will be minimal.

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Code:

class Solution {
    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;
    }
}

来源：CSDN

作者：Boston_Kimisong

链接：https://blog.csdn.net/weixin_45744426/article/details/103708510