找出数组中最小的k个数。
- 排序 ,$nlog(n)$
- 维持大小为k的最大堆,$nlog(k)$
- 快排思想,期望时间复杂度 $O(n)$,最坏时间复杂度 $O(n^2)$,随机partition会改善最坏的情况
class Solution { public: int partition(vector<int>& nums, int l, int r) { int pivot = nums[r]; int i = l - 1; for (int j = l; j < r; ++j) { if (nums[j] < pivot) { i++; swap(nums[i], nums[j]); } } swap(nums[i + 1], nums[r]); return i + 1; } int randomized_partition(vector<int>& nums, int l, int r) { int i = rand() % (r - l + 1) + l; swap(nums[i], nums[r]); return partition(nums, l, r); } void randomized_selected(vector<int>& nums, int l, int r, int k) { if (l >= r) return; int pos = randomized_partition(nums, l, r); int num = pos - l + 1; if (k == num) return; else if (k < num) randomized_selected(nums, l, pos - 1, k); else randomized_selected(nums, pos + 1, r, k - num); } vector<int> getLeastNumbers(vector<int>& arr, int k) { srand((unsigned)time(NULL)); randomized_selected(arr, 0, (int)arr.size() - 1, k); vector<int> res(arr.begin(), arr.begin() + k); return res; } };