python
1冒泡法
找出最小的k个数输出(超出时间限制)
class Solution: def smallestK(self, arr: List[int], k: int) -> List[int]: count = 0 for i in range(0,len(arr)): min = i for j in range(i,len(arr)): if(arr[j]
2.直接用python自带的函数
class Solution: def smallestK(self, arr: List[int], k: int) -> List[int]: arr.sort() return arr[0:k]