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:
- n is a positive integer, which is in the range of [1, 10000].
- All the integers in the array will be in the range of [-10000, 10000].
题目标签:Array
这道题目给了我们一个数组有2n integers, 需要我们把这个数组分成n对,然后从每一对里面拿小的那个数字,把所有的加起来,返回这个sum。并且要使这个sum 尽量最大。如何让sum 最大化呢,我们想一下,如果是两个数字,一个很小,一个很大,这样的话,取一个小的数字,就浪费了那个大的数字。所以我们要使每一对的两个数字尽可能接近。我们先把nums sort 一下,让它从小到大排列,接着每次把index: 0, 2, 4...偶数位的数字加起来就可以了。
Java Solution:
Runtime beats 83.27%
完成日期:05/10/2017
关键词:Array
关键点:Sort
public class Solution
{
public int arrayPairSum(int[] nums)
{
int sum = 0; Arrays.sort(nums); for(int i=0; i<nums.length; i+=2)
sum += nums[i]; return sum;
}
}
参考资料:N/A
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