最近刷题，有篇博客将经典的排序算法都讲得挺好，但原文是Java实现。
特此将我对应实现的Python版贴出如下～
博客链接如下：https://blog.csdn.net/weixin_41190227/article/details/86600821
- Compare Sort 比较排序
- Bubble Sort 冒泡排序, Select Sort 选择排序, Insertion Sort 插入排序, Shell Sort 希尔排序, Merge Sort 归并排序, Quick Sort 快速排序, Heap Sort 堆排序,
- Non-compare Sort 非比较排序
- Counting Sort 计数排序, Bucket Sort 桶排序, Raidx Sort 基数排序
def bubble_sort(arr):
"""
Time complexity:
avg: O(n^2); best: O(n); worst: O(n^2)
note: "best: O(n)" while use swapTag
Space complexity:
O(1)
"""
if len(arr) == 0:
return arr
for i in range(len(arr)):
tag = True
for j in range(len(arr) - i - 1):
if arr[j] > arr[j + 1]:
arr[j], arr[j + 1] = arr[j + 1], arr[j]
tag = False
if tag:
break
return arr
def select_sort(arr, tag):
"""
Time complexity:
avg: O(n^2); best: O(n^2); worst: O(n^2)
Space complexity:
O(1)
表现稳定，适合数据规模较小的情况
"""
if len(arr) == 0:
return arr
def min_select_sort(arr):
for i in range(len(arr)):
min_idx = i
for j in range(i + 1, len(arr)):
if arr[min_idx] > arr[j]:
min_idx = j
arr[i], arr[min_idx] = arr[min_idx], arr[i]
return arr
def max_select_sort(arr):
for i in range(len(arr) - 1, -1, -1):
max_idx = i
for j in range(i):
if arr[max_idx] < arr[j]:
max_idx = j
arr[i], arr[max_idx] = arr[max_idx], arr[i]
return arr
if tag:
return min_select_sort(arr)
else:
return max_select_sort(arr)
def insertion_sort(arr):
"""
Time complexity:
avg: O(n^2); best: O(n); worst: O(n^2)
Space complexity:
O(1)
表现稳定，适合数据规模较小的情况
"""
if len(arr) == 0:
return arr
for i in range(len(arr) - 1):
tmp, j = arr[i + 1], i
while j >= 0 and tmp < arr[j]:
arr[j + 1] = arr[j]
j -= 1
arr[j + 1] = tmp
return arr
def shell_sort(arr):
"""
Time complexity:
avg: O(nlgn); best: O(nlgn); worst: O(nlgn)
Space complexity:
O(1)
缩小增量的插入排序
"""
if len(arr) == 0:
return arr
gap = len(arr) // 2
while gap > 0:
for i in range(gap, len(arr)):
tmp, idx = arr[i], i - gap
while idx >= 0 and arr[idx] > tmp:
arr[idx + gap] = arr[idx]
idx -= gap
arr[idx + gap] = tmp
gap //= 2
return arr
def merge_sort(arr):
"""
Time complexity:
avg: O(nlgn); best: O(nlgn); worst: O(nlgn)
Space complexity:
O(n)
"""
if len(arr) < 2:
return arr
def merge(left, right):
tmp, l, r, li, ri = [], len(left), len(right), 0, 0
for idx in range(l + r):
if li >= l or (li < l and ri < r and left[li] > right[ri]):
tmp.append(right[ri])
ri += 1
else:
tmp.append(left[li])
li += 1
return tmp
mid = len(arr) // 2
left_arr = merge_sort(arr[:mid])
right_arr = merge_sort(arr[mid:])
return merge(left_arr, right_arr)
def quick_sort(arr):
"""
choose a pivot as base
sort list as two parts: smaller than pivot, bigger than pivot
recursive sort the left part and right part
Time complexity:
avg: O(nlgn); best: O(nlgn); worst: O(n^2)
Space complexity:
O(nlgn) or O(lgn) : O(1*lgn) here as O(1) for partition and O(lgn) for recursion times
"""
def recursion(ar, start, end):
if len(ar) == 0 or start < 0 or end >= len(ar) or start > end:
return []
idx = partition(ar, start, end)
if idx > start:
recursion(ar, start, idx - 1)
if idx < end:
recursion(ar, idx + 1, end)
return ar
def partition(ar, start, end):
from random import randint
pivot = randint(start, end)
idx = start - 1 # the last position of smaller part
ar[pivot], ar[end] = ar[end], ar[pivot]
for i in range(start, end + 1):
if ar[i] <= ar[end]: # use <= to put the pivot to the middle of partition
idx += 1
ar[i], ar[idx] = ar[idx], ar[i]
# # if use "ar[i] < ar[end]"
# idx += 1
# ar[idx], ar[end] = ar[end], ar[idx]
return idx
return recursion(arr, 0, len(arr) - 1)
def heap_sort(arr):
"""
build a standard max heap
each round: swap the max heap(first element) to expand the sorted part, and adjust the left part
Time complexity:
avg: O(nlgn); best: O(nlgn); worst: O(nlgn)
Space complexity:
O(lg1): in place
"""
if len(arr) < 2:
return arr
def adjust(ar, i, l):
# l is the total length of unsorted partition
idx = i
for j in (i * 2, i * 2 + 1): # if left or right child is bigger
if j < l and ar[j] > ar[idx]:
idx = j
if idx != i: # if any swap, adjust the sub tree to max heap
ar[idx], ar[i] = ar[i], ar[idx]
adjust(ar, idx, l)
# build max heap first by adjusting each sub tree as heap
for i in range(len(arr) // 2 - 1, -1, -1):
adjust(arr, i, len(arr))
# each round: move the heap to sorted right partition, and adjust left unsorted partition
for i in range(len(arr) - 1, 0, -1):
arr[0], arr[i] = arr[i], arr[0]
adjust(arr, 0, i)
return arr
def counting_sort(arr):
"""
find the min and max num of array
count the occurrences of i, store into the i'th place of tmp array
fill the original array according to counts
note: here use bias to reduce the extra space of tmp array
Time complexity:
avg: O(n+k); best: O(n+k); worst: O(n+k)
k = max_num - min_num while use bias
Space complexity:
O(k)
"""
if len(arr) < 2:
return arr
min_n = max_n = arr[0]
for n in arr: # find the min and max num
if n < min_n:
min_n = n
if n > max_n:
max_n = n
bias, counts = 0 - min_n, [0] * (max_n - min_n + 1) # use bias to reduce extra space
for n in arr:
counts[n + bias] += 1
idx = 0
for n, count in enumerate(counts):
if count != 0:
arr[idx:idx + count] = [n - bias] * count
idx += count
return arr
def bucket_sort(arr, bucket_size):
"""
upper level of counting sort
put nums to buckets according to nums' value mapping
use bucket sort recursively or other sort method to sort each bucket
factually, bucket_size is the gap of values instead size of buckets ——> bucket_size==1, but bucket may be [1,1,1]
Time complexity:
avg: O(n+k); best: O(n+k); worst: O(n^2)
k = max_num - min_num while use bias
Space complexity:
O(n+k)
"""
if len(arr) < 2:
return arr
min_n = max_n = arr[0]
for n in arr:
if n < min_n:
min_n = n
if n > max_n:
max_n = n
bucket_count, buckets = (max_n - min_n) // bucket_size + 1, []
for i in range(bucket_count):
buckets.append([])
for n in arr: # map num to buckets as back buckets have larger numbers
buckets[(n - min_n) // bucket_size].append(n)
ans = []
for bucket in buckets:
if bucket_size == 1: # do not use "ans.append(bucket[0])" as the bucket may have repeated nums
ans += bucket
else:
if bucket_count == 1: # to avoid endless loop in one bucket
bucket_size -= 1
tmp = bucket_sort(bucket, bucket_size)
ans += tmp
return ans
def raidx_sort(arr):
"""
sort from low digit to high digit
Time complexity:
avg: O(n*k); best: O(n*k); worst: O(n*k)
k = max_num - min_num while use bias
Space complexity:
O(n+k)
"""
if len(arr) < 2:
return arr
max_n = max(arr)
max_digit, buckets, mod, div = len(str(max_n)), [], 10, 1
for i in range(10):
buckets.append([])
for i in range(max_digit):
for n in arr:
buckets[(n % mod) // div].append(n)
arr = []
for bucket in buckets:
for n in bucket:
arr.append(n)
bucket.clear()
mod, div = mod * 10, div * 10
return arr
# TODO: doubt the time complexity of Bucket Sort and Raidx Sort
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