17. Letter Combinations of a Phone Number
Given a string containing digits from 2-9
inclusive, 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. Note that 1 does not map to any letters.
Example:
Input: "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.
题意
题目给你一串2-9的数字, 需要返回所有可能的键盘字母组合,映射规则就是手机九键输入的键盘映射规则
这道题可以用递归的方法也可以不用递归的方法
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如果不用递归的方法,可以用Python列表生成式的来表示这种排列组合的关系
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也可以用递归的方法,每次递归确定一位数字,当全部都确定后递归结束
解法
- 递归
class Solution(object):
def letterCombinations(self, digits):
"""
:type digits: str
:rtype: List[str]
"""
dic = {2: ['a', 'b', 'c'],
3: ['d', 'e', 'f'],
4: ['g', 'h', 'i'],
5: ['j', 'k', 'l'],
6: ['m', 'n', 'o'],
7: ['p', 'q', 'r', 's'],
8: ['t', 'u', 'v'],
9: ['w', 'x', 'y', 'z'],
}
ret_str = []
if len(digits) == 0: return []
# 递归出口,当递归到最后一个数的时候result拿到结果进行for循环遍历
if len(digits) == 1:
return dic[int(digits[0])]
result = self.letterCombinations(digits[1:])
for r in result:
for j in dic[int(digits[0])]:
ret_str.append(j + r)
return ret_str
- 非递归
class Solution:
def letterCombinations(self, digits):
"""
:type digits: str
:rtype: List[str]
"""
if not digits:
return []
digit2chars = {
'2': 'abc',
'3': 'def',
'4': 'ghi',
'5': 'jkl',
'6': 'mno',
'7': 'pqrs',
'8': 'tuv',
'9': 'wxyz'
}
res = [ i for i in digit2chars[digits[0]] ]
for i in digits[1:]:
res = [ m+n for m in res for n in digit2chars[i] ]
return res
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