此题与<<矩阵中的路径>>那一题为同一类问题,方法是采用回溯法来解决,我同样参照剑指offer官方的解释进行了代码实现,思路可以看我<<矩阵中的路径>>的那一题,他们是很相似的哦!!!
如果您看到我的代码有什么问题或者需要改进的请您指正哦!!!!!
class Solution:
def movingCount(self, threshold, rows, cols):
# write code here
if threshold < 0:
return 0
visited = [0]*cols*rows
result = self.movingCountCore(threshold, 0, 0 ,rows, cols, visited)
del visited
return result
def movingCountCore(self, threshold, row, col ,rows, cols, visited):
count = 0
if self.check(threshold, row, col ,rows, cols, visited):
visited[row * cols + col] = 1
count = 1 + self.movingCountCore(threshold, row+1, col ,rows, cols, visited)+\
self.movingCountCore(threshold, row-1, col ,rows, cols, visited)+\
self.movingCountCore(threshold, row, col+1 ,rows, cols, visited)+\
self.movingCountCore(threshold, row, col-1 ,rows, cols, visited)
return count
def check(self, threshold, row, col ,rows, cols, visited):
if (row >= 0 and row < rows and col >=0 and col < cols and (not visited[row * cols + col]) and ((self.getDigitSum(row)+self.getDigitSum(col))<=threshold)):
return True
return False
def getDigitSum(self, number):
sum_ = 0
for i in str(number):
sum_ += int(i)
return sum_

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