Prove safety offer40. Top K

Find the minimum number k of the array.

1. Sort, $ nlog (n) $
2. Maintaining the size of the largest heap of k, $ nlog (k) $
3. Quick drain thought desirable time complexity $ O (n) $, worst case time complexity $ O (n ^ 2) $, random partition improves the worst case 

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;
    }
};