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









