leetcode 561---- ArrayPartition

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].

题意 ，给你一个2n长度数组，分成n个长度为2的数组，使得这n个数组的最小值总和最小

            简答题目， 把所有的数存入一个排序数组中，每2个数取一个最小值就可以了

  

int arrayPairSum(int* nums,int numsSize) {    int i = 0,n=numsSize;    int sum = 0;    int store[20001];    int flag = 0;    for(i=0;i<20001;i++){        store[i]=0;    }    for(i=0;i<n;i++){             store[nums[i]+10000]++;    }    for(i=0;i<20001;i++){        if(store[i]>0){        if(flag == 0){            sum+= (i-10000);            store[i]--;            flag = 1;            if(store[i]>0){                i--;            }        }else{            if(store[i]>0){                store[i]--;            }            if(store[i]>0){                i--;            }            flag = 0;        }        }else{            continue;        }    }    return sum;}