题目：原题链接（中等）
标签：字符串
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N ) | O ( N ) | 84ms (72.55%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Codec:
def encode(self, strs: [str]) -> str:
lst = [str(len(s)) for s in strs]
return ",".join(lst) + "." + "".join(strs)
def decode(self, s: str) -> [str]:
idx = s.index(".")
code = s[:idx]
if not code:
return []
lst = [int(ss) for ss in code.split(",")]
idx += 1
ans = []
for length in lst:
ans.append(s[idx:idx + length])
idx += length
return ans
