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 caseif (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;
}