Array Partition I

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].

方法1：排序，然后从第一个数字开始，每隔一个数字进行相加。

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

网上新代码：
方法2：空间换时间，速度快，占用空间大。

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;
    }
};