leetcode-array-561-array-partition-i
用java好好刷点题吧。。。从简单到复杂。 题目： https://leetcode.com/problems/array-partition-i/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:
- 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].
题目大意： 给2n个数字，把他们两个一组，分成n组。 使得每一组的最小值之和最大。 求最大值。 解： 每组都是取的最小值，损失了另外一个数，为了使损失最小，另外一个数最好是离取得的数最近的数。即将两个相近的数分为一组。。 排序后，就满足需求。对奇数位的数求和。
public class Solution {
public int arrayPairSum(int\[\] nums) {
Arrays.sort(nums);
int result = 0;
for (int i = 0; i < nums.length; i += 2) {
result += nums\[i\];
}
return result;
}
}