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 (a<b)
return 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;
}
};