LeetCode Array Partition I

原题链接在这里：https://leetcode.com/problems/array-partition-i/#/description

题目：

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].

题解：

排序之后，相邻两个分到一起能保证每组最小值相加最大.

Time Complexity: O(nlogn). n = nums.length. Space: O(1).

AC Java:

 1 public class Solution {
 2     public int arrayPairSum(int[] nums) {
 3         if(nums == null || nums.length == 0){
 4             return 0;
 5         }
 6         
 7         Arrays.sort(nums);
 8         int res = 0;
 9         for(int i = 0; i<nums.length; i+=2){
10             res += nums[i];
11         }
12         return res;
13     }
14 }

LeetCode Array Partition I

原文：http://www.cnblogs.com/Dylan-Java-NYC/p/7076179.html