LeetCode 561. Array Partition I

题目描述：
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:
n is a positive integer, which is in the range of [1, 10000].
All the integers in the array will be in the range of [-10000, 10000].

解题思路：
2n个数字分成n组，每组2个数字，取每组中较小的数字，求和。要求是得到的和最大，方法返回值即为这个最大和。

设2n个数字升序排序后为：a0，a1，a2，a3，….，a(2n-1)。
猜想一下容易发现，最大和即为 (a0+a2+a4+….+a(2n-2))，最小和即为(a0+a1+a2+…+a(n-1))。
得出这个结论后，我们马上就可以编码实现啦。

AC代码：

class Solution {
public:
    int getMin(int a, int b) {
        if (areturn a;
        else
            return b;
}

    int arrayPairSum(vector<int> nums) {
        int sum = 0;
        sort(nums.begin(), nums.end());
        for (int i = 0; i<(int)nums.size(); i = i + 2) {
            sum = sum + getMin(nums[i], nums[i + 1]);
        }
        return sum;
    }
};