561. Array Partition I

Description :

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 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].

 解法 :
          這題要求我們把array想像成一個pair的群，然後求每個pair的最小值並加總起來的和
，求出最大者。
          其實很簡單，先將這個array進行排序這樣就可以保證，每個pair的最左邊數字一定是最小，而且下一個pair的最小值又會比上一個pair的最大值大。

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
    
        int pairmax = 0;
        
        sort(nums.begin(), nums.end());
        
        for(int i = 0; i < nums.size(); i+=2)
        {
            pairmax += nums[i]; 
        }
        
        return pairmax;
    }
};