561. Array Partition I(C语言)

这道题什么鬼

不应该输出5，[2,3],一组，[1,4]一组么

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.

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].

这道题我又理解错题意了，还好柏宁晚饭时候，帮我讲了讲

这个min(ai,bi)的意思是（ai，bi）这个数对里最小的值ai或者bi，比如[1,3]里是1,[2,4]里是2

sum of min（ai，bi） from i to n的意思的是，所有1到n这n个数对里，相对较小的值的和

所以题目要求是：

1.求数对里的最小值

2.对所有数对的最小值求和

3.使这个和最大

解题方法就是：2n个数，从小到大排序，然后奇数下标相加

写了一个冒泡排序，超时，改成了快排，AC

int arrayPairSum(int* nums, int numsSize) {    int i=0,j=numsSize-1;    int sum=0;        void sort(int i,int j,int* nums);//函数声明    sort(0,numsSize-1, nums);    for(i=0;i<numsSize;i=i+2){        sum+=nums[i];    }        return sum;}void sort(int i,int j,int* nums){    int low=i;    int high=j;    int key=nums[i];    int temp=0;        if(i>j){        return ;    }    while(i<j){        while(nums[j]>key&&i<j){            j--;        }        nums[i]=nums[j];        while(nums[i]<=key&&i<j){            i++;                    }        nums[j]=nums[i];            }        nums[i]=key;    sort(low,i-1,nums);    sort(i+1,high,nums);    }