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:
- 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].
class Solution {
public:
void quicksort(vector<int>& nums, int start, int end){//快排！！！
int i = start;
int j = end;
int x = nums[i];
if (start < end){
while (i<j)
{
while (nums[j]<x && i<j)
j--;
if (i<j)
nums[i++] = nums[j];
while (nums[i]>x && i<j)
i++;
if (i<j)
nums[j--] = nums[i];
}
nums[i] = x;
quicksort(nums, start, i-);
quicksort(nums, i+, end);
}
}
public:
int arrayPairSum(vector<int>& nums) {
//直接冒泡排序是超时的，所以考虑选择一个效率更高的排序
int len = nums.size();
quicksort(nums,,len-);
//贪心策略：排序选择偶数下标，相加
int sum = ;
for (int i=;i<len;i+=){
sum += nums[i];
}
return sum;
}
};