手撕代码必会排序算法(冒泡、快排、堆排)
一、冒泡排序
1、基础版
def bubble_Sort(arr):
n = len(arr)
for i in range(n-1):
for j in range(n-i-1):
if arr[j] > arr[j+1]:
arr[j], arr[j+1] = arr[j+1], arr[j]
2、改进版
加入Flag,若剩余的都已经是有序的则无需继续
def super_bubble_Sort(arr):
n = len(arr)
for i in range(n-1):
Flag = False
for j in range(n-i-1):
if arr[j] > arr[j+1]:
arr[j], arr[j+1] = arr[j+1], arr[j]
Flag = True
if not Flag: break
二、快排
1、方法一
思路及其简单,实现也及其简单,但是面试官往往想让你写的是方法二
def quick_Sort(nums):
if len(nums) <= 1:
return nums
pivot = nums[0]
left = [nums[i] for i in range(1, len(nums)) if nums[i]<pivot]
right = [nums[i] for i in range(1, len(nums)) if nums[i]>=pivot]
return quick_Sort(left) + [pivot] + quick_Sort(right)
2、方法二
import random
def partition(left, right, nums):
rand_index = random.randint(left, right)
nums[left], nums[rand_index] = nums[rand_index], nums[left]
pivot, index = nums[left], left
for i in range(left+1, right+1):
if nums[i] < pivot:
index += 1
nums[i], nums[index] = nums[index], nums[i]
nums[left], nums[index] = nums[index], nums[left]
return index
def quick_Sort_Inter(left, right, nums):
if left < right:
index = partition(left, right, nums)
quick_Sort_Inter(left, index-1, nums)
quick_Sort_Inter(index+1, right, nums)
三、堆排
def heap_Sort(nums):
#调整堆
def Heapify(nums, i, size):
#非叶子节点的左右两个孩子
lchild = 2*i + 1
rchild = 2*i + 2
#在当前节点、左右孩子中找到最大元素的索引
largest = i
if lchild < size and nums[lchild] > nums[largest]:
largest = lchild
if rchild < size and nums[rchild] > nums[largest]:
largest = rchild
#如果最大元素的索引不是当前节点,把最大的节点交换到上面,继续调整堆
if largest != i:
nums[largest], nums[i] = nums[i], nums[largest]
Heapify(nums, largest, size)
#建立堆
def build_Heap(nums, size):
for i in range(len(nums)//2)[::-1]:
Heapify(nums, i, size)
size = len(nums)
build_Heap(nums, size)
for i in range(size)[::-1]:
#每次根节点都是最大的数,最大数放在后面
nums[0], nums[i] = nums[i], nums[0]
#交换完后还需要继续调整堆,只需调整根节点,此时数组的size不包括已经排序好的数
Heapify(nums, 0, i)
return nums
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