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。
示例:
输入："23" 输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
第一次理解: 输入“234”, 组合可以为2个字母也可以为3个字母。
思路: 求出所有的2个字母的组合，在2个字母组合的基础上加入第3个
def letterCombinations(digits):
"""
:type digits: str
:rtype: List[str]
"""
# 数字到对应字母集的映射.比如2 对应[a,b,c]
num2char = {}
i1 = 2
num = 97
while num < 123:
num2char[i1] = []
c = 0
b = 3
if i1 == 7 or i1 == 9:
b = b + 1
while c < b:
num2char[i1].append(chr(num))
num = num + 1
c = c + 1
i1 = i1 + 1
nums = []
for ch in digits:
nums.append(int(ch))
result = []
result1 = []
for s1 in range(0, len(nums)-1, 1):
for s2 in range(s1+1, len(nums), 1):
result.append([s1, s2])
tmp = []
for c1 in num2char[nums[s1]]:
for c2 in num2char[nums[s2]]:
tmp.append(c1+c2)
result1.append(tmp)
s = 0
e = len(result)
while s < e:
c = 0
for i in range(s, e, 1):
l = result[i]
t = result1[i]
for j in range(l[-1] + 1, len(nums), 1):
tmp = []
l1 = l.copy()
l1.append(j)
result.append(l1)
for f1 in t:
for f2 in num2char[nums[j]]:
tmp.append(f1+f2)
result1.append(tmp)
c = c + 1
s = e
e = e + c
print(result)
print(len(result))
print(result1)
print(len(result1))
letterCombinations("2345")第二次理解: 题意的意思是：输入3个数字，那么输出必须为3个字母的组合。 不能少于3个
python代码：
class Solution:
def letterCombinations(self, digits):
"""
:type digits: str
:rtype: List[str]
"""
if len(digits) == 0:
return []
# 数字到对应字母集的映射.比如2 对应[a,b,c]
num2char = {}
i1 = 2
num = 97
while num < 123:
num2char[i1] = []
c = 0
b = 3
if i1 == 7 or i1 == 9:
b = b + 1
while c < b:
num2char[i1].append(chr(num))
num = num + 1
c = c + 1
i1 = i1 + 1
nums = []
for ch in digits:
nums.append(int(ch))
if len(nums) == 1:
return num2char[nums[0]]
result = num2char[nums[0]]
for num in nums[1:]:
tmp = []
for c1 in result:
for c2 in num2char[num]:
tmp.append(c1 + c2)
result = tmp
return result