力扣坑题：存在重复元素

int getkey(int* a, int left, int right)
{
	int mid = (left + right) / 2;
	if (a[left] > a[right])
	{
		if (a[mid] < a[right])
		{
			return right;
		}
		else if (a[mid] < a[left])
		{
			return mid;
		}
		else
		{
			return left;
		}
	}
	else
	{
		if (a[mid] < a[left])
		{
			return left;
		}
		else if (a[mid] < a[right])
		{
			return mid;
		}
		else
		{
			return right;
		}
	}
}
 
 
void fastsort(int* a, int begin, int end)
{
	if (begin >= end)
	{
		return;
	}
	int left = begin;
	int right = end;
	int cur = left + 1;
	int mid = getkey(a, left, right);
	int c = a[left];
	a[left] = a[mid];
	a[mid] = c;
	int key = a[left];
	while (cur <= right)
	{
		if (a[cur] < key)
		{
			int d = a[left];
			a[left] = a[cur];
			a[cur] = d;
			left++;
			cur++;
		}
		else if (a[cur] > key)
		{
			int e = a[right];
			a[right] = a[cur];
			a[cur] = e;
			--right;
		}
		else
		{
			cur++;
		}
	}
	fastsort(a, begin, left - 1);
	fastsort(a, right + 1, end);
}


bool containsDuplicate(int* nums, int numsSize){
    fastsort(nums,0,numsSize-1);
    for(int i=1;i<numsSize;i++)
    {
        if(nums[i]==nums[i-1])
        {
           return true;
        }
    }
    return false;
}

本题有巨量数据和针对快排的数据，所以要用三路分治秒