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Array Partition I

时间:2017-08-24 09:51:02      阅读:311      评论:0      收藏:0      [点我收藏+]

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].

 

分析:将给出的整型数组分成两两一组后,返回每组最小值之和,要求和最大。

思路:分组时,让每组的差的平方和最小即可。最简单的方式,直接排序,依次分组。

 JAVA CODE

class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int sum = 0;
        for(int i = 0; i < nums.length; i += 2){
            sum += Math.min(nums[i],nums[i+1]);
        }
        return sum;
    }
}

 

 

Array Partition I

原文:http://www.cnblogs.com/baichangfu/p/7421207.html

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