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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.
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].
把2n个数分成n组,找出这n组数中最小的数相加,使得最小数的总和尽可能的大
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; }
561. Array Partition I
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