前段时间腾讯视频面遇到的题目，第一个方法用冒泡实现时间复杂度太高，又用堆优化了一遍，还可以进一步优化，下面总结一下
面试题029 最小的K个数
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,。
解题思路：排序
代码 速度很快
class Solution(object):
def smallestK(self, arr, k):
"""
:type arr: List[int]
:type k: int
:rtype: List[int]
"""
if not arr or len(arr)<k:
return []
arr.sort()
return arr[:k]
代码 From leetcode 速度慢
# 标准快排
class Solution(object):
def smallestK(self, arr, k):
"""
:type arr: List[int]
:type k: int
:rtype: List[int]
"""
def quick_sort(nums):
#用2个数组来分别保存跟基准相比大或者小的数
left,right = [],[]
if(len(nums) < 2):
return nums
else:
base = nums[0]
#每一次以数组的第一个值为基准，小于等于的放到左边数组，大于的放到右边数组
left = [num for num in nums[1:] if num <= base]
right = [num for num in nums[1:] if num > base]
#如果刚好左边的小数组序列长度符合要求就可以直接返回了，因为题目没有要求返回顺序
if(len(left) == k):
return left
else:
return quick_sort(left) + [base] + quick_sort(right)
return quick_sort(arr)[:k]
# 归并排序
class Solution:
def GetLeastNumbers_Solution(self, tinput, k):
# write code here
def merge_sort(lst):
if len(lst) <= 1:
return lst
mid = len(lst) // 2
left = merge_sort(lst[: mid])
right = merge_sort(lst[mid:])
return merge(left, right)
def merge(left, right):
l, r, res = 0, 0, []
while l < len(left) and r < len(right):
if left[l] <= right[r]:
res.append(left[l])
l += 1
else:
res.append(right[r])
r += 1
res += left[l:]
res += right[r:]
return res
if tinput == [] or k > len(tinput):
return []
tinput = merge_sort(tinput)
return tinput[: k]
# 堆排序
class Solution:
def GetLeastNumbers_Solution(self, tinput, k):
# write code here
def siftup(lst, temp, begin, end):
if lst == []:
return []
i, j = begin, begin * 2 + 1
while j < end:
if j + 1 < end and lst[j + 1] > lst[j]:
j += 1
elif temp > lst[j]:
break
else:
lst[i] = lst[j]
i, j = j, 2 * j + 1
lst[i] = temp
def heap_sort(lst):
if lst == []:
return []
end = len(lst)
for i in range((end // 2) - 1, -1, -1):
siftup(lst, lst[i], i, end)
for i in range(end - 1, 0, -1):
temp = lst[i]
lst[i] = lst[0]
siftup(lst, temp, 0, i)
return lst
if tinput == [] or k > len(tinput):
return []
tinput = heap_sort(tinput)
return tinput[: k]
# 冒泡排序
class Solution:
def GetLeastNumbers_Solution(self, tinput, k):
# write code here
def bubble_sort(lst):
if lst == []:
return []
for i in range(len(lst)):
for j in range(1, len(lst) - i):
if lst[j-1] > lst[j]:
lst[j-1], lst[j] = lst[j], lst[j-1]
return lst
if tinput == [] or k > len(tinput):
return []
tinput = bubble_sort(tinput)
return tinput[: k]
# 直接选择
class Solution:
def GetLeastNumbers_Solution(self, tinput, k):
# write code here
def select_sort(lst):
if lst == []:
return []
for i in range(len(lst)-1):
smallest = i
for j in range(i, len(lst)):
if lst[j] < lst[smallest]:
smallest = j
lst[i], lst[smallest] = lst[smallest], lst[i]
return lst
if tinput == [] or k > len(tinput):
return []
tinput = select_sort(tinput)
return tinput[: k]
# 插入排序
class Solution:
def GetLeastNumbers_Solution(self, tinput, k):
# write code here
def Insert_sort(lst):
if lst == []:
return []
for i in range(1, len(lst)):
temp = lst[i]
j = i
while j > 0 and temp < lst[j - 1]:
lst[j] = lst[j - 1]
j -= 1
lst[j] = temp
return lst
if tinput == [] or k > len(tinput):
return []
tinput = Insert_sort(tinput)
return tinput[: k]