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本文实例讲述了Python实现查找最小的k个数。分享给大家供大家参考,具体如下:

题目描述
输入n个整数,找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字,则最小的4个数字是1,2,3,4,。
解法1
使用partition函数可以知道,使用==O(N)==的时间复杂度就可以找出第K大的数字,并且左边的数字比这个数小,右边的数字比这个数字大。因此可以取k为4,然后输出前k个数字,如果需要排序的话再对结果进行排序
# -*- coding:utf-8 -*-
class Solution:
def PartitionOfK(self, numbers, start, end, k):
if k < 0 or numbers == [] or start < 0 or end >= len(numbers) or k > end:
return
low, high = start, end
key = numbers[low]
while low < high:
while low < high and numbers[high] >= key:
high -= 1
numbers[low] = numbers[high]
while low < high and numbers[low] <= key:
low += 1
numbers[high] = numbers[low]
numbers[low] = key
if low < k:
self.PartitionOfK(numbers, start + 1, end, k)
elif low > k:
self.PartitionOfK(numbers, start, end - 1, k)
def GetLeastNumbers_Solution(self, tinput, k):
# write code here
if k <= 0 or tinput == [] or k > len(tinput):
return []
self.PartitionOfK(tinput, 0, len(tinput) - 1, k)
return sorted(tinput[0:k])
#测试:
sol = Solution()
listNum = [4,5,1,6,2,7,3,8]
rel = sol.GetLeastNumbers_Solution(listNum, 4)
print(rel)