文章作者：Tyan 博客：noahsnail.com | CSDN | 简书
1. Description
2. Solution
**解析：**Version 1，先分离数字和字母，如果二者数量差的绝对值大于1，则不可能形成字母数字字符串，否则，根据数量多少，按顺序依次连接字符数字数组即可。
- Version 1
class Solution:
def reformat(self, s: str) -> str:
letters = []
digits = []
for ch in s:
if ch.isalpha():
letters.append(ch)
else:
digits.append(ch)
diff = len(letters) - len(digits)
if abs(diff) > 1:
return ''
elif diff == 0 or diff == 1:
arr1 = letters
arr2 = digits
else:
arr1 = digits
arr2 = letters
result = [0] * len(s)
for i in range(len(arr1)):
result[i * 2] = arr1[i]
for i in range(len(arr2)):
result[i * 2 1] = arr2[i]
return ''.join(result)**解析：**Version 2，进一步优化。
- Version 2
class Solution:
def reformat(self, s: str) -> str:
letters = 0
digits = 0
for ch in s:
if ch.isalpha():
letters = 1
else:
digits = 1
diff = letters - digits
if abs(diff) > 1:
return ''
elif diff == 0 or diff == 1:
i = 0
j = 1
else:
i = 1
j = 0
result = [0] * len(s)
for ch in s:
if ch.isalpha():
result[i] = ch
i = 2
else:
result[j] = ch
j = 2
return ''.join(result)**解析：**Version 2，使用双指针，分别从两端找起，如果碰到字母和数字不符合当前位置，则交换二者。
- Version 3
class Solution:
def reformat(self, s: str) -> str:
letters = 0
digits = 0
for ch in s:
if ch.isalpha():
letters = 1
else:
digits = 1
diff = letters - digits
if abs(diff) > 1:
return ''
elif diff == 0 or diff == 1:
i = 0
j = len(s) - 2
else:
i = 1
j = len(s) - 1
result = list(s)
while i < len(s) and j >=0:
while i < len(s) and result[i].isalpha():
i = 2
while j >=0 and result[j].isdigit():
j -= 2
if i < len(s) and j >=0:
result[i], result[j] = result[j], result[i]
return ''.join(result)Reference
- https://leetcode.com/problems/reformat-the-string/
