剑指 Offer 40. 最小的k个数https://leetcode.cn/problems/zui-xiao-de-kge-shu-lcof/

法1：二叉堆

通过最小堆，直接筛选出最小的k个数

vector<int> getLeastNumbers(vector<int>& arr, int k) {

	priority_queue<int, vector<int>, greater<int>> minHeap;

	for (const int num : arr)
	{
		minHeap.push(num);
	}

	vector<int> res(k);
	for (int i = 0; i < k; ++i)
	{
		res[i] = minHeap.top();
		minHeap.pop();
	}

	return res;
}

法2：快速选择

二分+快排 结合

将最小的k个数 分到一边 

int partition(vector<int>& nums, const int left, const int right)
{
	// base case
	if (left >= right)
	{
		return left;
	}

	int pivot = nums[left];

	int i = left + 1, j = right;
	while (i <= j)
	{
		while (i <= right && nums[i] <= pivot)
		{
			++i;
		}
		while (j > left && nums[j] > pivot)
		{
			--j;
		}

		if (i > j)
		{
			break;
		}

		swap(nums[i], nums[j]);
	}
	swap(nums[left], nums[j]);

	return j;
}

vector<int> getLeastNumbers(vector<int>& arr, int k) {

	int n = arr.size();

	int left = 0, right = n - 1;
	while (left <= right)
	{
		int mid = partition(arr, left, right);
		if (mid < k - 1)
		{
			left = mid + 1;
		}
		else if (mid > k - 1)
		{
			right = mid - 1;
		}
		else
		{
			break;
		}
	}

	vector<int> res(k);
	for (int i = 0; i < k; ++i)
	{
		res[i] = arr[i];
	}

	return res;
}