Python实现排序算法
2017-11-27 本文已影响0人
rogeroyer
快速排序
print(sorted([1, 0, 9, 2, 3, 6, 5], reverse=False)) # python list 排序函数 #
def quick_sort(lists, left, right):
if left >= right:
return lists
key = lists[left]
low = left
high = right
while left < right:
while left < right and lists[right] >= key:
right -= 1
lists[left] = lists[right]
while left < right and lists[left] <= key:
left += 1
lists[right] = lists[left]
lists[right] = key
quick_sort(lists, low, left - 1)
quick_sort(lists, left + 1, high)
return lists
插入排序
def insert_sort(lists):
count = len(lists)
for i in range(1, count):
key = lists[i]
j = i - 1
while j >= 0:
if lists[j] > key:
lists[j + 1] = lists[j]
lists[j] = key
j -= 1
return lists
归并排序
def merge(left, right):
i, j = 0, 0
result = []
while i < len(left) and j < len(right):
if left[i] <= right[j]:
result.append(left[i])
i += 1
else:
result.append(right[j])
j += 1
result += left[i:]
result += right[j:]
return result
def merge_sort(lists):
if len(lists) <= 1:
return lists
num = len(lists) / 2
left = merge_sort(lists[:num])
right = merge_sort(lists[num:])
return merge(left, right)
希尔排序
def shell_sort(lists):
count = len(lists)
step = 2
group = count / step
while group > 0:
for i in range(0, group):
j = i + group
while j < count:
k = j - group
key = lists[j]
while k >= 0:
if lists[k] > key:
lists[k + group] = lists[k]
lists[k] = key
k -= group
j += group
group /= step
return lists