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Python full stack road series recursion

高洛峰
高洛峰Original
2017-02-09 10:44:161242browse

The so-called recursion actually means that the function itself calls the function until the specified conditions are met and then exits the function layer by layer. For example,

Once upon a time, there was a mountain, and there was a temple in the mountain. There was an old monk in the temple, who was giving the child The monk is telling stories! What's the story? "Once upon a time, there was a mountain, and there was a temple in the mountain. There was an old monk in the temple, and he was telling a story to the young monk! What is the story? 'Once upon a time, there was a mountain, and there was a temple in the mountain. There was an old monk in the temple, and he was telling a story to the young monk. Tell a story to the little monk! What is the story?...'"

  • ##Use functions to write a Fibonacci sequence

##0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368

The Fibonacci sequence is to add the two numbers in front to get the next number, and then proceed in sequence

The code is as follows
#!/usr/bin/env python
# _*_ coding: utf-8 _*_

def Counter(n1, n2):
    if n1 > 10000:  # 当要计算的值大于10000就退出
        return
    print("Counter:", n1)  # 输出当前计算到那个值了
    n3 = n1 + n2  # 第一个值加上第一个值等于第三个值
    Counter(n2, n3)  # 调用计数器函数,此时第一个值是调用函数传过来的最后一个值,而第二个值是计算出来的第三个值


Counter(0, 1)  # 调用计数器函数

Output results

/usr/bin/python3.5 /home/ansheng/Documents/PycharmProjects/blogcodes/斐波那契.py
Counter: 0
Counter: 1
Counter: 1
Counter: 2
Counter: 3
Counter: 5
Counter: 8
Counter: 13
Counter: 21
Counter: 34
Counter: 55
Counter: 89
Counter: 144
Counter: 233
Counter: 377
Counter: 610
Counter: 987
Counter: 1597
Counter: 2584
Counter: 4181
Counter: 6765

Process finished with exit code 0

    Use recursion to obtain the 10th number in the Fibonacci sequence and return the value to the caller
  • Code:
#!/usr/bin/env python
# _*_ coding: utf-8 _*_

def Counter(Index, Start, End):
    print("第%d次计算,第一个数字是%d,第二个数字是%d" % (Index, Start, End))
    if Index == 10:  # 如果要计算的值是10就退出
        return Start
    N = Start + End  # N等于第一个数加上第二个数
    Number = Counter(Index + 1, End, N)  # 继续调用计数器函数,End相当与传给函数的第一个数,N是传给函数的第二个数
    return Number


result = Counter(1, 0, 1)
print("得出的数字是:", result)

Output result

/usr/bin/python3.5 /home/ansheng/Documents/PycharmProjects/blogcodes/递归.py
第1次计算,第一个数字是0,第二个数字是1
第2次计算,第一个数字是1,第二个数字是1
第3次计算,第一个数字是1,第二个数字是2
第4次计算,第一个数字是2,第二个数字是3
第5次计算,第一个数字是3,第二个数字是5
第6次计算,第一个数字是5,第二个数字是8
第7次计算,第一个数字是8,第二个数字是13
第8次计算,第一个数字是13,第二个数字是21
第9次计算,第一个数字是21,第二个数字是34
第10次计算,第一个数字是34,第二个数字是55
得出的数字是: 34

Process finished with exit code 0

Original link


The so-called recursion is actually the function itself calling the function until Exit functions layer by layer after meeting the specified conditions. For example,

Once upon a time there was a mountain, a temple in the mountain, and an old monk in the temple who was telling stories to the young monk! What's the story? "Once upon a time, there was a mountain, and there was a temple in the mountain. There was an old monk in the temple, and he was telling a story to the young monk! What is the story? 'Once upon a time, there was a mountain, and there was a temple in the mountain. There was an old monk in the temple, and he was telling a story to the young monk. Tell a story to the little monk! What is the story?...'"

##Use functions to write a Fibonacci sequence
  • ##0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368

The Fibonacci sequence is to add the two numbers in front to get the next number, and then proceed in sequence

The code is as follows

#!/usr/bin/env python
# _*_ coding: utf-8 _*_

def Counter(n1, n2):
    if n1 > 10000:  # 当要计算的值大于10000就退出
        return
    print("Counter:", n1)  # 输出当前计算到那个值了
    n3 = n1 + n2  # 第一个值加上第一个值等于第三个值
    Counter(n2, n3)  # 调用计数器函数,此时第一个值是调用函数传过来的最后一个值,而第二个值是计算出来的第三个值


Counter(0, 1)  # 调用计数器函数
Output results
/usr/bin/python3.5 /home/ansheng/Documents/PycharmProjects/blogcodes/斐波那契.py
Counter: 0
Counter: 1
Counter: 1
Counter: 2
Counter: 3
Counter: 5
Counter: 8
Counter: 13
Counter: 21
Counter: 34
Counter: 55
Counter: 89
Counter: 144
Counter: 233
Counter: 377
Counter: 610
Counter: 987
Counter: 1597
Counter: 2584
Counter: 4181
Counter: 6765

Process finished with exit code 0

Use recursion to obtain the 10th number in the Fibonacci sequence and return the value to the caller

  • Code:

    #!/usr/bin/env python
    # _*_ coding: utf-8 _*_
    
    def Counter(Index, Start, End):
        print("第%d次计算,第一个数字是%d,第二个数字是%d" % (Index, Start, End))
        if Index == 10:  # 如果要计算的值是10就退出
            return Start
        N = Start + End  # N等于第一个数加上第二个数
        Number = Counter(Index + 1, End, N)  # 继续调用计数器函数,End相当与传给函数的第一个数,N是传给函数的第二个数
        return Number
    
    
    result = Counter(1, 0, 1)
    print("得出的数字是:", result)
  • Output result
/usr/bin/python3.5 /home/ansheng/Documents/PycharmProjects/blogcodes/递归.py
第1次计算,第一个数字是0,第二个数字是1
第2次计算,第一个数字是1,第二个数字是1
第3次计算,第一个数字是1,第二个数字是2
第4次计算,第一个数字是2,第二个数字是3
第5次计算,第一个数字是3,第二个数字是5
第6次计算,第一个数字是5,第二个数字是8
第7次计算,第一个数字是8,第二个数字是13
第8次计算,第一个数字是13,第二个数字是21
第9次计算,第一个数字是21,第二个数字是34
第10次计算,第一个数字是34,第二个数字是55
得出的数字是: 34

Process finished with exit code 0

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