题解 | 无序数组中最小的k个数

class KthNumbers {
public:
    vector<int> findKthNumbers(vector<int>& A, int n, int k) {
        if (k <= 0 || n <= 0) return vector<int>();
        
        // 复制数组并排序
        vector<int> sorted = A;
        sort(sorted.begin(), sorted.end());
        
        // 获取第k小的数
        int kthValue = sorted[k - 1];
        
        // 遍历原数组，保留小于等于kthValue的前k个数
        vector<int> result;
        for (int i = 0; i < n && result.size() < k; i++) {
            if (A[i] <= kthValue) {
                result.push_back(A[i]);
            }
        }
        
        return result;
    }
};

import java.util.*;

public class KthNumbers {
    public int[] findKthNumbers(int[] A, int n, int k) {
        if (k <= 0 || n <= 0) return new int[0];
        
        // 复制数组并排序
        int[] sorted = A.clone();
        Arrays.sort(sorted);
        
        // 获取第k小的数
        int kthValue = sorted[k - 1];
        
        // 遍历原数组，保留小于等于kthValue的前k个数
        int[] result = new int[k];
        int index = 0;
        
        for (int i = 0; i < n && index < k; i++) {
            if (A[i] <= kthValue) {
                result[index++] = A[i];
            }
        }
        
        return result;
    }
}

class KthNumbers:
    def findKthNumbers(self, A, n, k):
        if k <= 0 or n <= 0:
            return []
            
        sorted_arr = sorted(A)
        kth_value = sorted_arr[k - 1]
        
        result = []
        for num in A:
            if len(result) < k and num <= kth_value:
                result.append(num)
                
        return result