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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