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].
这道题让我们分割数组,两两一对,让每对中较小的数的和最大。这题难度不大,用贪婪算法就可以了。由于我们要最大化每对中的较小值之和,那么肯定是每对中两个数字大小越接近越好,因为如果差距过大,而我们只取较小的数字,那么大数字就浪费掉了。明白了这一点,我们只需要给数组排个序,然后按顺序的每两个就是一对,我们取出每对中的第一个数即为较小值累加起来即可,参见代码如下:
class Solution {
public:
int arrayPairSum(vector<int>& nums) {
int res = , n = nums.size();
sort(nums.begin(), nums.end());
for (int i = ; i < n; i += ) {
res += nums[i];
}
return res;
}
};