[Leetcode]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].

思路：对于每一对数 min (a,b)，我们如果选中了一个 a ，那么对于另一个数b 一定要满足 b-a 最小。例如，1 2 3 4这四个数。

组成的对应该是(1,2),(3,4)。而不是(1,3),(2,4)。所以具体做法就是排序，然后选奇数位上的数相加就好了。代码如下：

 

1 class Solution {
2     public int arrayPairSum(int[] nums) {
3         Arrays.sort(nums);
4         int sum = 0;
5         for (int i=0;i<nums.length-1;i+=2)
6             sum+=nums[i];
7         return sum;
8     }
9 }

 

转载于:https://www.cnblogs.com/David-Lin/p/7735034.html