LeetCode 561.Array Partition I

LeetCode 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), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

Example:

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).

分析：

这道题很简单，当然不考虑时间复杂度的情况下，我们直接给vector按从小到大排序，然后从小到大每两个数取最小的那一个，显然加起来就是结果。

代码如下：

#include <iostream>#include <vector>#include <algorithm>using namespace std;class Solution {public:    int arrayPairSum(vector<int>& nums) {        sort(nums.begin(), nums.end());        int n = nums.size();        int sum = 0;        for (int i = 0; i < n; i += 2) {            cout << min(nums[i], nums[i + 1]) << endl;            sum += min(nums[i], nums[i + 1]);        }        return sum;    }};int main() {    Solution s;    int n;    cin >> n;    vector<int> nums(n * 2);    for (int i = 0; i < n * 2; i++) {        cin >> nums[i];    }    cout << s.arrayPairSum(nums) << endl;    return 0;}