刷题 两数之和

https://leetcode.cn/problems/two-sum/submissions/588870256/?envType=study-plan-v2&envId=top-100-liked
参考快排算法 https://blog.csdn.net/oSKyTonight/article/details/129813861
/**

• Note: The returned array must be malloced, assume caller calls free().
  */
  int Swap(int *a, int *b)
  {
  int c = *a;
  *a = *b;
  *b = c;
  return 0;
  }

int quickSort(int *a, int *b, int left, int right)
{
if (left >= right)
return 0;
int begin = left, end = right;

int keyi = left;//选定序列首元素为基准值

while (left < right)
{
	//右边找小
	while (left < right && a[right] >= a[keyi])
		right--;
	//左边找大
	while (left < right && a[left] <= a[keyi])
		left++;
	//交换
	Swap(&a[left], &a[right]);
    Swap(&b[left], &b[right]);
}
//相遇时交换key和相遇位置的值
Swap(&a[keyi], &a[left]);
Swap(&b[keyi], &b[left]);

keyi = left;//用keyi记录下本轮确定的基准值位置

quickSort(a, b, begin, keyi - 1);//递归排序左区间[begin , keyi-1]
quickSort(a, b, keyi + 1, end);//递归排序右区间[keyi+1 , end]
return 0;

}

int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
int *retList = (int *)malloc(2 * sizeof(int));
int *tempOrder = (int *)malloc(numsSize * sizeof(int));
*returnSize = 2;
memset(retList, 0, 2 * sizeof(int));
int a = 0;
int b = numsSize-1;

for (int i = 0; i < numsSize; i++)
{
    tempOrder[i] = i;
}
quickSort(nums, tempOrder, 0, numsSize-1);
while(a<b)
{
    if (nums[a]+nums[b]==target)
    {
        retList[0] = tempOrder[a];
        retList[1] = tempOrder[b];
        break;
    }
    else if(nums[a] + nums[b] < target)
    {
        a++;

    }
    else
    {
        b--;
    }
}
free(tempOrder);
return retList;

}