561. Array Partition I

标签：max   input   oss   []   .so   pairs   turn   from   put   

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:

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].

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

标签：max   input   oss   []   .so   pairs   turn   from   put   

原文地址：http://www.cnblogs.com/sunli0205/p/6773142.html