@author: ZZQ
@software: PyCharm
@file: letterCombinations.py
@time: 2018/10/18 18:33
要求:给定一个仅包含数字 2-9 的字符串,返回所有它能表示的字母组合。
给出数字到字母的映射如下(与电话按键相同)。注意 1 不对应任何字母。
2:abc; 3: def; 4:ghi; 5: jkl; 6: mno; 7: pqrs; 8: tuv; 9: wxyz
输入:"23"
输出:["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
思路:先得出每隔数字代表的字母,然后两两合并再合并。(记录奇偶)
class Solution():
def __init__(self):
pass
def numToLetter(self, num):
dict = {2: 'abc',
3: 'def',
4: "ghi",
5: "jkl",
6: "mno",
7: "pqrs",
8: "tuv",
9: "wxyz"
}
return dict[num]
def combine(self, str1, str2):
len1 = len(str1)
len2 = len(str2)
combined = []
temp_str = ''
for i in range(len1):
for j in range(len2):
combined.append(temp_str+str1[i]+str2[j])
temp_str = ''
return combined
def letterCombinations(self, digits):
"""
:type digits: str
:rtype: List[str]
"""
if digits == "":
return []
length = len(digits)
letters = []
for i in range(length):
cur_num = int(digits[i])
cur_letter = self.numToLetter(cur_num)
letters.append(cur_letter)
if len(letters) == 1:
return [letters[0][t] for t in range(len(letters[0]))]
while len(letters) > 1:
cur_letters = []
if len(letters) % 2:
for i in range((len(letters) - 1) / 2):
cur_letters.append(self.combine(letters[i * 2], letters[i * 2 + 1]))
cur_letters.append(letters[len(letters)-1])
else:
for i in range(len(letters) / 2):
cur_letters.append(self.combine(letters[i * 2], letters[i * 2 + 1]))
letters = cur_letters
return letters[0]
if __name__ == "__main__":
answer = Solution()
print answer.letterCombinations('2')
# print len(answer.letterCombinations('2345'))