LeetCode

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: 4Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

Note:

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

我的做法是把数组排了个序，然后每次取相邻的两个数求最小值，时间复杂度O(nlogn)，空间复杂度O(1)

看评论区有O(n)的做法，用数组存每个数有多少个，然后遍历，依然每次取两个数中较小的那个。空间复杂度O(n)

class Solution {public:    int arrayPairSum(vector<int>& nums) {        int ans = 0;        sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size(); i += 2) {            ans += min(nums[i], nums[i+1]);        }        return ans;    }};

class Solution {public:    int arrayPairSum(vector<int>& nums) {        vector<int> hashtable(20001,0);        for(size_t i=0;i<nums.size();i++)        {            hashtable[nums[i]+10000]++;        }        int ret=0;        int flag=0;        for(size_t i=0;i<20001;){            if((hashtable[i]>0)&&(flag==0)){                ret=ret+i-10000;                flag=1;                hashtable[i]--;            }else if((hashtable[i]>0)&&(flag==1)){                hashtable[i]--;                flag=0;            }else i++;        }        return ret;    }};