561. Array Partition I

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].

大致意思是给你一个数组，两两一组，把这两个数的最小值加起来，求这个和的最大值

先上代码吧

import java.util.Arrays;
class Solution {
    public int arrayPairSum(int[] nums) {
        int result = 0;
        Arrays.sort(nums);
        for(int i=0;i<nums.length;i=i+2)
        {
            result = result + nums[i];
        }
        return result;
    }
}

思路是这样的，首先一个数组的和是给定的我们定位sum

sum=a1+a2+...+an+b1+b2+...+bn

假设所有的b都比a大 差值表示为di=bi-ai

则sum=2（a1+a2+..+an）+d1+d2+...+dn

那么问题就变为了求最小的d1+d2+...+dn

显然相邻的两个数组成一组，di最小

那么我们只要把数组从小到大排序，然后取所有下标为偶数的求和既可