leetcode 561. Array Partition I（C语言）10

贴原题：

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 =

min(1, 2) + min(3, 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].

解析

  本题我用了刚刚做另一道题时http://blog.csdn.net/m0_37454852/article/details/78062299 写的快速排序算法，就很简单了，排好序之后每隔一个相加就行了。

贴代码：

void my_qsort(int* nums, int l, int r)
{
if(l<r)
{
int i=l+1;//左端点，从基准值的下一个开始比较
int j=r;//右端点
while(i<j)
{
if(*(nums+i)>*(nums+l))//若比关键值大，则把该值放到右端点（交换该值与右端点值）
{
int temp=*(nums+j);
*(nums+j)=*(nums+i);
*(nums+i)=temp;
j--;//右端点右移
}
else//否则就选择下一个继续比较
{
i++;
}
}
if(*(nums+i)>=*(nums+l))
{
i--;
}
int temp=*(nums+i);
*(nums+i)=*(nums+l);
*(nums+l)=temp;
my_qsort(nums, l, i-1);//递归调用
my_qsort(nums, i+1, r);
}
}
int arrayPairSum(int* nums, int numsSize) {
my_qsort(nums, 0, numsSize-1);
int sum=0;
for(int i=0; i<numsSize; i+=2)
{
sum+=*(nums+i);
}
return sum;
}