递归
作者：互联网
递归
递归的三定律
- 要有基本情况，作为终止条件
- 必须改变状态并向基本情况靠近
- 以递归的方式调用自身
1、将整数转换为任意进制的字符串
def toStr(num, base):
convertString = "0123456789ABCDEF"
if num < base:
return convertString[num]
else:
return toStr(num//base, base) + convertString[num%base]
print(toStr(10,8))
print(toStr(10,2))
print(toStr(78,16))
12
1010
4E
2、螺旋矩
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,200)
myWin.exitonclick()
- 图1
3、分形树
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()
- 图2
3、锡尔宾斯基三角
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()
4、河内塔问题
def moveTower(height,fromPole, toPole, withPole):
if height >= 1:
moveTower(height-1,fromPole,withPole,toPole)
moveDisk(fromPole,toPole)
moveTower(height-1,withPole,toPole,fromPole)
def moveDisk(fp,tp):
print("moving disk from",fp,"to",tp)
moveTower(2,"A","B","C")
moving disk from A to C
moving disk from A to B
moving disk from C to B
5、探索迷宫
import turtle
PART_OF_PATH = 'O'
TRIED = '.'
OBSTACLE = '+'
DEAD_END = '-'
class Maze:
def __init__(self,mazeFileName):
rowsInMaze = 0
columnsInMaze = 0
self.mazelist = []
mazeFile = open(mazeFileName,'r')
rowsInMaze = 0
for line in mazeFile:
rowList = []
col = 0
for ch in line[:-1]:
rowList.append(ch)
if ch == 'S':
self.startRow = rowsInMaze
self.startCol = col
col = col + 1
rowsInMaze = rowsInMaze + 1
self.mazelist.append(rowList)
columnsInMaze = len(rowList)
self.rowsInMaze = rowsInMaze
self.columnsInMaze = columnsInMaze
self.xTranslate = -columnsInMaze/2
self.yTranslate = rowsInMaze/2
self.t = turtle.Turtle()
self.t.shape('turtle')
self.wn = turtle.Screen()
self.wn.setworldcoordinates(-(columnsInMaze-1)/2-.5,-(rowsInMaze-1)/2-.5,(columnsInMaze-1)/2+.5,(rowsInMaze-1)/2+.5)
def drawMaze(self):
self.t.speed(10)
self.wn.tracer(0)
for y in range(self.rowsInMaze):
for x in range(self.columnsInMaze):
if self.mazelist[y][x] == OBSTACLE:
self.drawCenteredBox(x+self.xTranslate,-y+self.yTranslate,'orange')
self.t.color('black')
self.t.fillcolor('blue')
self.wn.update()
self.wn.tracer(1)
def drawCenteredBox(self,x,y,color):
self.t.up()
self.t.goto(x-.5,y-.5)
self.t.color(color)
self.t.fillcolor(color)
self.t.setheading(90)
self.t.down()
self.t.begin_fill()
for i in range(4):
self.t.forward(1)
self.t.right(90)
self.t.end_fill()
def moveTurtle(self,x,y):
self.t.up()
self.t.setheading(self.t.towards(x+self.xTranslate,-y+self.yTranslate))
self.t.goto(x+self.xTranslate,-y+self.yTranslate)
def dropBreadcrumb(self,color):
self.t.dot(10,color)
def updatePosition(self,row,col,val=None):
if val:
self.mazelist[row][col] = val
self.moveTurtle(col,row)
if val == PART_OF_PATH:
color = 'green'
elif val == OBSTACLE:
color = 'red'
elif val == TRIED:
color = 'black'
elif val == DEAD_END:
color = 'red'
else:
color = None
if color:
self.dropBreadcrumb(color)
def isExit(self,row,col):
return (row == 0 or
row == self.rowsInMaze-1 or
col == 0 or
col == self.columnsInMaze-1 )
def __getitem__(self,idx):
return self.mazelist[idx]
def searchFrom(maze, startRow, startColumn):
# try each of four directions from this point until we find a way out.
# base Case return values:
# 1. We have run into an obstacle, return false
maze.updatePosition(startRow, startColumn)
if maze[startRow][startColumn] == OBSTACLE :
return False
# 2. We have found a square that has already been explored
if maze[startRow][startColumn] == TRIED or maze[startRow][startColumn] == DEAD_END:
return False
# 3. We have found 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
myMaze = Maze('maze2.txt')
myMaze.drawMaze()
myMaze.updatePosition(myMaze.startRow,myMaze.startCol)
searchFrom(myMaze, myMaze.startRow, myMaze.startCol)
True
6、动态规划
硬币找零
def recMC(coinValueList,change):
minCoins = change # 全部使用1美分兑换为硬币最多的情形
if change in coinValueList:
return 1
else:
for i in [c for c in coinValueList if c <= change]:
numCoins = 1 + recMC(coinValueList,change-i)
if numCoins < minCoins:
minCoins = numCoins
return minCoins
print(recMC([1,5,10,25],63))
6
- 上述算法非常低效，原因在于有太多的重复计算
#性能更好
def recDC(coinValueList,change,knownResults):
minCoins = change
if change in coinValueList:
knownResults[change] = 1
return 1
elif knownResults[change] > 0:
return knownResults[change]
else:
for i in [c for c in coinValueList if c <= change]:
numCoins = 1 + recDC(coinValueList, change-i,
knownResults)
if numCoins < minCoins:
minCoins = numCoins
knownResults[change] = minCoins
return minCoins
print(recDC([1,5,10,25],63,[0]*64))
6
- 上面的算法并不是真正的动态规划算法，下面使用真正的动态规划
def dpMakeChange(coinValueList,change,minCoins,coinsUsed):
for cents in range(change+1):
coinCount = cents
newCoin = 1
for j in [c for c in coinValueList if c <= cents]:
if minCoins[cents-j] + 1 < coinCount:
coinCount = minCoins[cents-j]+1
newCoin = j
minCoins[cents] = coinCount
coinsUsed[cents] = newCoin
return minCoins[change]
def printCoins(coinsUsed,change):
coin = change
while coin > 0:
thisCoin = coinsUsed[coin]
print(thisCoin)
coin = coin - thisCoin
def main():
amnt = 63
clist = [1,5,10,21,25]
coinsUsed = [0]*(amnt+1)
coinCount = [0]*(amnt+1)
print("Making change for",amnt,"requires：")
print(dpMakeChange(clist,amnt,coinCount,coinsUsed),"coins")
print("They are:")
printCoins(coinsUsed,amnt)
print("The used list is as follows:")
print(coinsUsed)
main()
Making change for 63 requires：
3 coins
They are:
21
21
21
The used list is as follows:
[1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 10, 1, 1, 1, 1, 5, 1, 1, 1, 1, 10, 21, 1, 1, 1, 25, 1, 1, 1, 1, 5, 10, 1, 1, 1, 10, 1, 1, 1, 1, 5, 10, 21, 1, 1, 10, 21, 1, 1, 1, 25, 1, 10, 1, 1, 5, 10, 1, 1, 1, 10, 1, 10, 21]
标签：递归,color,self,startRow,points,myTurtle,def 来源： https://blog.csdn.net/FANGLICHAOLIUJIE/article/details/89358862