DAY8：leetcode #17 Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

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class Solution(object):
def addPointer(self,pointer_l,pointer_lf):
for i in range(len(pointer_l))[::-1]:
if pointer_l[i] != pointer_lf[i]:
pointer_l[i] += 1
for j in range(i+1,len(pointer_l)):
pointer_l[j] = 0
return pointer_l
return False
def letterCombinations(self, digits):
"""
:type digits: str
:rtype: List[str]
"""
if digits == '':
return []
map_dict = {
'2':'abc',
'3':'def',
'4':'ghi',
'5':'jkl',
'6':'mno',
'7':'pqrs',
'8':'tuv',
'9':'wxyz'
}
letter_l = []
pointer_l = []
pointer_temp = []
result = []
for di in list(digits):
letter_l.append(list(map_dict[di]))
pointer_l.append(len(map_dict[di])-1)
pointer_temp.append(0)
while(True):
temp = ''
if not pointer_temp:
return result
for i in range(len(pointer_l)):
temp += letter_l[i][pointer_temp[i]]
result.append(temp)
pointer_temp = self.addPointer(pointer_temp,pointer_l)


看似简单，做了一些之后才发现思路有问题。花了挺长时间的。