数据结构与算法python版（2）-递归与汉诺塔问题

递归简介

• 递归的两个特点：调用自身和结束条件

如：

def func(x):
if x>0:
print(x)
func(x-1)

func(3)

执行结果如下：

3
2
1

这里需要注意一下，如将打印语句放到下面，如下代码，结果将是完全不一样的

def func(x):
if x>0:
func(x-1)
print(x)

func(3)

执行结果如下：

1
2
3

2、汉诺塔问题

• 汉诺塔问题：



• 当n=2时操作步骤：
  原始状况如下，目标是将A上的两个都移动到C上



• 步骤一：将A上的小盘从A移动到B



• 步骤二：将A盘上的大盘移动到C



• 步骤三：将B上的小盘移动到C，结束



• 当n个盘时，思路是将A上的n-1个盘看做是一个小盘，将A最下面的大盘看做是一个大盘，此时即和n=2时的思路一致了



• 步骤一：将A上n-1个盘移动到B



• 步骤二：将A上最下面的大盘移动到C盘



• 步骤三：将B上的n-1个盘移动到C盘上
• 算法实现如下：

def hanoi(n,a,b,c):
"""
将n个盘从a经过b移动到c
:param n:
:param a:
:param b:
:param c:
:return:
"""
if n>0:
hanoi(n-1,a,c,b)
print(f"moving {n} from {a} to {c}")
hanoi(n-1,b,a,c)

if __name__=="__main__":
print("---------------- n=2 -----------------------------")
hanoi(2,"A","B","C")
print("---------------- n=3 -----------------------------")
hanoi(3, "A", "B", "C")
print("---------------- n=4 -----------------------------")
hanoi(4, "A", "B", "C")

执行结果如下：

---------------- n=2 -----------------------------
moving 1 from A to B
moving 2 from A to C
moving 1 from B to C
---------------- n=3 -----------------------------
moving 1 from A to C
moving 2 from A to B
moving 1 from C to B
moving 3 from A to C
moving 1 from B to A
moving 2 from B to C
moving 1 from A to C
---------------- n=4 -----------------------------
moving 1 from A to B
moving 2 from A to C
moving 1 from B to C
moving 3 from A to B
moving 1 from C to A
moving 2 from C to B
moving 1 from A to B
moving 4 from A to C
moving 1 from B to C
moving 2 from B to A
moving 1 from C to A
moving 3 from B to C
moving 1 from A to B
moving 2 from A to C
moving 1 from