算法-----python实现

Posted 2021-01-15 hnlmy

斐波那契数列

def f(n):
    if n == 1:
        return 1
    elif n == 2:
        return 1
    else:
        return f(n-1)+f(n-2)

print(f(8))

用普通函数实现斐波那契数列：

def f(n):
    li = [0,1,1]
    if n <=2:
        return li[n]
    for i in range(3,n+1):
        li.append(li[-1]+li[-2])
    return li[n]

print(f(8))

常见的时间复杂度（按照效率排序）

O(1)<O(logn)<O(n)<O(nlogn)<O(n2)<O(n2logn)<O(n3)

递归实例：汉诺塔问题

def hanoi(n,A,B,C):
    if n > 0:
        hanoi(n-1,A,C,B)
        print("%s->%s"%(A,C))
        hanoi(n-1,B,A,C)

hanoi(4,"A","B","C")

二分查找

def binary_search(li, val):
    low = 0
    high = len(li) - 1
    while low <= high:
        mid = (low + high) // 2
        if li[mid] < val:
            low = mid + 1
        elif li[mid] > val:
            high = mid -1
        else:
            return mid
    return None

冒泡排序

import random


def bubble_sort(li):
    for i in range(len(li)-1):  # i表示第i趟,共n-1趟
        # 第i趟 无序区范围 0~n-i-1
        for j in range(len(li)-i-1):
            if li[j] > li[j+1]:
                li[j],li[j+1] = li[j+1],li[j]
        print(li)


li = [8,5,7,9,4,2,6,1,3]
bubble_sort(li)

冒泡排序------优化

@cal_time
def bubble_sort_2(li):
    for i in range(len(li)-1):
        exchange = False
        for j in range(len(li)-1):
            if li[j] > li[j+1]:
                li[j],li[j+1] = li[j+1],li[j]
                exchange = True
        if not exchange:
            return

li=list(range(10000))
random.shuffle(li)
bubble_sort_2(li)