[LeetCode] 561.Array Partition I 备忘

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]. 

数组分离。在一个偶数长度的数组中，每2个数字组成一组，计算每一组中的最小值的和的最大值。

第一眼看见时，就想到数组排序，然后两两组队，就行了。

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

虽然成功了但是耗时86ms，去评论区看了下，发现一个巧妙的方法，跟桶排序非常相似。由于题目已经确定所有数字的值在-10000到10000之间，于是定义一个维度为20001的int数组，数组下标索引就想到于该题中输入数组nums中每个数字，即作为键，类似于map，该数组中每个值用来统计题中输入数组中的每个数字出现的次数，只不过下标的值多加了10000，就自动将所有数字排序了，然后每隔一个数字取一次与总值累加即可，耗时62ms。

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