数据结构｜python实现简单数据结构

Recursion

def fact(n):
    if n == 1:
        return 1
    else:
        return n*fact(n-1)
a = fact(5)
print(a)
## This is a Recursion

Binary Search

## linear search -> bruteforce
##linear search first
def sqrt(p):
    for i in range(p):
        if i * i == p:
            return i
a = sqrt(25)
##Here is binary search
def sqrt2(n):
    low  = 0
    high = n
    while low <= high:
        mid =  (low + high) // 2
        if mid * mid == n:
            return mid
        if mid * mid > n:
            high = mid - 1
        else:
            low = mid + 1
a = sqrt2(25)
print(a)

Print fibnacci numbers in 3 ways

##ordinary method
def f1(n):
    if n == 0:
        return 0
    elif n == 1 or n == 2:
        return 1
    else:
        return f1(n-1) + f1(n-2)

##using dictionary(with notebook)
def f2(n, nb = {}):
    ##look up f2(n) in notebook first
    if n in nb:
        return nb[n]
    if n == 0:
        return 0
    if n == 1:
        return 1
    #if not in notebook, then put it in
    ans = f2(n-1, nb) + f2(n-2, nb)
    nb[n] = ans
    return ans

##using iteration
def f3(n):
    a = 0
    b = 1
    for i in range(n):
        a, b = b, a+b
    return a

print(f1(9))
print(f2(9))
print(f3(9))

Stack in python

##stack
##First in last out
##plates = [ ]
##plate.append(1) -> contain [1]
##plate.append(2) -> contain [1, 2]
##x = plate.pop() -> x = 2
##plate.append(4) -> [1,4]
##y = plate.pop() -> y = 4
##len(plates) = 1
def rev(nums):
    ans = []
    while len(nums) > 0:
        x = nums.pop()
        ans.append(x)
    return ans
nums = [1, 2, 4, 6, 8]
y = rev(nums)
print(y)

Queue in python

##Queue
##First in first out
##using the list in python like stack
##stack = [] queue = []
q = []
q.append(1)
q.append(2)
q.append(5)
q.append(14)
##q[-1] -> last item
##peek -> q[0] while queue
##pop -> pop(0) pop first item
##peek -> q[-1] while stack
##pop -> pop() pop last item
def sqr(nums):
    ans= []
    while len(nums) > 0:
        x = nums.pop(0)
        ##if want to reverse:
        ## x = nums.pop( )
        ##              *
        ans.append(x*x)
    return ans

print(sqr(q))