数据结构python实现（二）递归

（二）递归

1. 什么样的问题能够用递归来解决

递归需要满足三个条件：

（1）一个问题的解可以分解成几个子问题的解

（2）这个问题与分解之后的子问题，除了数据规模不同，求解思路完全一样	

（3）存在递归终止条件

写递归代码的关键也是，找到递推公式和终止条件

2. 案例

案例1 阶乘

#循环实现
#时间O(n)
#空间O(1)
def factorial(n):
    result = 1
    for i in range(1, n+1):
        result *= i
    return result
  
#递归实现
#时间O(n)
#空间O(n)
ef factorial_recursive(n):
    if n == 1:
        return 1
    return n * factorial_recursive(n - 1)

案例2 斐波那契数列

#递归实现
#时间O(2^n)
def fibonacci1(n):
    assert(n>0)
    if (n <= 2): 
        return 1
    return fibonacci2(n-1) + fibonacci2(n-2)
  
  #循环实现
  #时间O(n)
  def fibonacci2(n):
    assert(n>0)
    a,b=0,1
    for i in range(1,n+1):
      a,b=b,a+b
    return a
  
  def fibonacci3(n):
    assert(n>0)
    if (n <= 1): 
        return (n,0)
    (a, b) = fibonacci3(n-1)
    return (a+b, a)

案例3 打印尺子

（热身）打印一下内容：

1

1 2 1

1 2 1 3 1 2 1

1 2 1 3 1 2 1 4 1 2 1 3 1 2 1

#时间O(2^n)
def ruler_bad(n):
    assert(n>=0)
    if (n==1):
        return "1"
    return ruler(n-1) + " " + str(n) + " " + ruler(n-1)

#时间O(n)
def ruler(n):
    assert(n>=0)
    if (n==1):
        return "1"
    t = ruler(n-1)
    return t + " " + str(n) + " " + t

#循环实现
def ruler2(n):
    result = ""
    for i in range(1, n+1):
        result = result + str(i) + " " + result
    return result

画尺子：

#画横线
def draw_line(tick_length, tick_label=‘‘):
    line = ‘-‘ * tick_length
    if tick_label:
        line += ‘ ‘ + tick_label
    print(line)
#画每一个大格子    
def draw_interval(center_length):
    if center_length > 0:
        draw_interval(center_length - 1)
        draw_line(center_length)
        draw_interval(center_length - 1)
#画多个格子    
def draw_rule(num_inches, major_length):
    draw_line(major_length, ‘0‘)
    for j in range(1, 1 + num_inches):
        draw_interval(major_length - 1)
        draw_line(major_length, str(j))

案例4 数字表达式

Given two integers a ≤ b, write a program that transforms a into b by a minimum sequence of increment (add 1) and unfolding (multiply by 2) operations.

For example,

23 = ((5 * 2 + 1) * 2 + 1)

113 = ((((11 + 1) + 1) + 1) * 2 * 2 * 2 + 1)

def intSeq(a, b):
    if (a == b):
        return str(a)
    
    if (b % 2 == 1):
        return "(" + intSeq(a, b-1) + " + 1)"
    
    if (b < a * 2):
        return "(" + intSeq(a, b-1) + " + 1)"
        
    return intSeq(a, b/2) + " * 2";

案例5 汉诺塔

def hanoi(n, start, end, by):
    if (n==1):
        print("Move from " + start + " to " + end)
    else:
        hanoi(n-1, start, by, end)
        hanoi(1, start, end, by)
        hanoi(n-1, by, end, start)

数据结构python实现（二）递归

原文：https://www.cnblogs.com/gdy1993/p/12984699.html