LeetCode刷题：561.Array Partition I

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), …, (a_n, b_n) which makes sum of min(a_i, b_i) for all i from 1 to n as large as possible.

Example:

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].

   解题思路：

   为得到最大的和，所以要保证第2大的数字和第1大的数字进行组合，第4大的数字和第3大的数字进行组合，以此类推。可以看出，我们将数组进行排序后，取2n+1项进行相加，即可得结果。

   解答：

   class Solution {
   public:
   int arrayPairSum(vector<int>& nums) {
       int len = nums.size();
       sort(nums.begin(), nums.end());
       int result = 0;
       for(int i = 0; i < len; ++i)
       {
           result += nums[i];
           ++i;
       }
       return result;
   }
   };