我試圖實(shí)現(xiàn)一個(gè)算法，該算法需要兩個(gè)in和n和k，其中n是連續(xù)的座位數(shù)，k是試圖坐在那一行的學(xué)生人數(shù)。問題在于每個(gè)學(xué)生在兩邊都必須至少有兩個(gè)座位。我有一個(gè)函數(shù)可以生成所有的子集(一個(gè)0或1的數(shù)組，1表示某人坐在那里)，然后我將它發(fā)送給一個(gè)函數(shù)來檢查它是否是一個(gè)有效的子集。這是我對該功能的代碼

def process(a,num,n):

c = a.count('1')

#If the number of students sitting down (1s) is equal to the number k, check the subset

if(c == num):

printa = True

for i in range(0,n):

if(a[i] == '1'):

if(i == 0):

if( (a[i+1] == '0') and (a[i+2] == '0') ):

break

else:

printa = False

elif(i == 1):

if( (a[i-1] == '0') and (a[i+1] == '0') and (a[i+2] == '0') ):

break

else:

printa = False

elif(i == (n-1)):

if( (a[i-2] == '0') and (a[i-1] == '0') and (a[i+1] == '0') ):

break

else:

printa = False

elif(i == n):

if( (a[i-2] == '0') and (a[i-1] == '0') ):

break

else:

printa = False

else:

if( (a[i-2] == '0') and (a[i-1] == '0') and (a[i+1] == '0') and (a[i+2] == '0') ):

break

else:

printa = False

if(printa):

print a

else:

return該代碼適用于k和n的小輸入，但如果我得到更高的值，出于某種原因我得不到列表錯(cuò)誤的索引。

任何幫助，非常感謝。

O輸入是一個(gè)看起來像這樣的列表

['1','0','0','1','0'] # a valid subset for n=5 and k=2

['0','0','0','1','1'] # an invalid subset編輯：

調(diào)用進(jìn)程的代碼：

'''

This function will recursivly call itself until it gets down to the leaves then sends that

subset to process function. It appends

either a 0 or 1 then calls itself

'''

def seatrec(arr,i,n,k):

if(i==n):

process(arr,k,n)

return

else:

arr.append("0")

seatrec(arr,i+1,n,k)

arr.pop()

arr.append("1")

seatrec(arr,i+1,n,k)

arr.pop()

return

'''

This is the starter function that sets up the recursive calls

'''

def seat(n,k):

q=[]

seat(q,0,n,k)

def main():

n=7

k=3

seat(n,k)

if __name__ == "__main__":

main()如果我使用這些數(shù)字，我得到的錯(cuò)誤是

if( (a[i-2] == '0') and (a[i-1] == '0') and (a[i+1] == '0') ):

IndexError: list index out of range

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