遞歸的三定律

遞歸算法必須具有基本情況。
遞歸算法必須改變其狀態并向基本情況靠近。
遞歸算法必須以遞歸方式調用自身。

整數轉換為任意進制字符串

def toStr(n,base):
   convertString = "0123456789ABCDEF"
   if n < base:
      return convertString[n]
   else:
      return toStr(n//base,base) + convertString[n%base]

print(toStr(1453,16))

棧幀：實現遞歸字符串轉換器

from pythonds.basic.stack import Stack

rStack = Stack()

def toStr(n,base):
    convertString = "0123456789ABCDEF"
    while n > 0:
        if n < base:
            rStack.push(convertString[n])
        else:
            rStack.push(convertString[n % base])
        n = n // base
    res = ""
    while not rStack.isEmpty():
        res = res + str(rStack.pop())
    return res

print(toStr(1453,16))

可視化遞歸

import turtle

myTurtle = turtle.Turtle()
myWin = turtle.Screen()

def drawSpiral(myTurtle, lineLen):
    if lineLen > 0:
        myTurtle.forward(lineLen)
        myTurtle.right(90)
        drawSpiral(myTurtle,lineLen-5)

drawSpiral(myTurtle,100)
myWin.exitonclick()

import turtle

def tree(branchLen,t):
    if branchLen > 5:
        t.forward(branchLen)
        t.right(20)
        tree(branchLen-15,t)
        t.left(40)
        tree(branchLen-15,t)
        t.right(20)
        t.backward(branchLen)

def main():
    t = turtle.Turtle()
    myWin = turtle.Screen()
    t.left(90)
    t.up()
    t.backward(100)
    t.down()
    t.color("green")
    tree(75,t)
    myWin.exitonclick()

main()

謝爾賓斯基三角形

import turtle

def drawTriangle(points,color,myTurtle):
    myTurtle.fillcolor(color)
    myTurtle.up()
    myTurtle.goto(points[0][0],points[0][1])
    myTurtle.down()
    myTurtle.begin_fill()
    myTurtle.goto(points[1][0],points[1][1])
    myTurtle.goto(points[2][0],points[2][1])
    myTurtle.goto(points[0][0],points[0][1])
    myTurtle.end_fill()

def getMid(p1,p2):
    return ( (p1[0]+p2[0]) / 2, (p1[1] + p2[1]) / 2)

def sierpinski(points,degree,myTurtle):
    colormap = ['blue','red','green','white','yellow',
                'violet','orange']
    drawTriangle(points,colormap[degree],myTurtle)
    if degree > 0:
        sierpinski([points[0],
                        getMid(points[0], points[1]),
                        getMid(points[0], points[2])],
                   degree-1, myTurtle)
        sierpinski([points[1],
                        getMid(points[0], points[1]),
                        getMid(points[1], points[2])],
                   degree-1, myTurtle)
        sierpinski([points[2],
                        getMid(points[2], points[1]),
                        getMid(points[0], points[2])],
                   degree-1, myTurtle)

def main():
   myTurtle = turtle.Turtle()
   myWin = turtle.Screen()
   myPoints = [[-100,-50],[0,100],[100,-50]]
   sierpinski(myPoints,3,myTurtle)
   myWin.exitonclick()

main()

漢諾塔游戲

def hanoi(n,x,y,z):
    if n==1:
        print(x,'-->',z)
    else:
        hanoi(n-1,x,z,y)#將前n-1個盤子從x移動到y上
        hanoi(1,x,y,z)#將最底下的最后一個盤子從x移動到z上
        hanoi(n-1,y,x,z)#將y上的n-1個盤子移動到z上
n=int(input('請輸入漢諾塔的層數：'))
hanoi(n,'x','y','z')

探索迷宮問題

def searchFrom(maze, startRow, startColumn):
    maze.updatePosition(startRow, startColumn)
   #  Check for base cases:
   #  1. We have run into an obstacle, return false
   if maze[startRow][startColumn] == OBSTACLE :
        return False
    #  2. We have found a square that has already been explored
    if maze[startRow][startColumn] == TRIED:
        return False
    # 3. Success, an outside edge not occupied by an obstacle
    if maze.isExit(startRow,startColumn):
        maze.updatePosition(startRow, startColumn, PART_OF_PATH)
        return True
    maze.updatePosition(startRow, startColumn, TRIED)

    # Otherwise, use logical short circuiting to try each
    # direction in turn (if needed)
    found = searchFrom(maze, startRow-1, startColumn) or 
            searchFrom(maze, startRow+1, startColumn) or 
            searchFrom(maze, startRow, startColumn-1) or 
            searchFrom(maze, startRow, startColumn+1)
    if found:
        maze.updatePosition(startRow, startColumn, PART_OF_PATH)
    else:
        maze.updatePosition(startRow, startColumn, DEAD_END)
    return found

參考：https://facert.gitbooks.io/python-data-structure-cn/4.%E9%80%92%E5%BD%92/4.11.%E6%8E%A2%E7%B4%A2%E8%BF%B7%E5%AE%AB/

總結

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