# Encode and Decode Strings

<https://leetcode.com/problems/encode-and-decode-strings/description/>

> Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.
>
> Machine 1 (sender) has the function:
>
> string encode(vector\<string> strs) {
>
> // ... your code
>
> return encoded\_string;
>
> }
>
> Machine 2 (receiver) has the function:
>
> vector\<string> decode(string s) {
>
> //... your code
>
> return strs;
>
> }
>
> So Machine 1 does:
>
> string encoded\_string = encode(strs);
>
> and Machine 2 does:
>
> vector\<string> strs2 = decode(encoded\_string);
>
> strs2 in Machine 2 should be the same as strs in Machine 1.
>
> Implement the encode and decode methods.
>
> Note:
>
> The string may contain any possible characters out of 256 valid ascii characters. Your algorithm should be generalized enough to work on any possible characters.
>
> Do not use class member/global/static variables to store states. Your encode and decode algorithms should be stateless.
>
> Do not rely on any library method such as eval or serialize methods. You should implement your own encode/decode algorithm.

## Thoughts

题目强烈暗示不能用单一字符作为分割符. 那么想记住string边界, 不难想到可以把长度记下来, 放在分隔符前. 然后每次读出时就把相应长度的substring抽出来放到结果中, index直接跳过它们进入下一个string的长度和分隔符.

注意只有长度没有分隔符也是不行的, 因为不知道数字合适结束.

## Code

```
public class Codec {

    // Encodes a list of strings to a single string.
    public String encode(List<String> strs) {
        StringBuilder sb = new StringBuilder();
        for (String str : strs) {
            sb.append(str.length()).append("/").append(str);
        }
        return sb.toString();
    }

    // Decodes a single string to a list of strings.
    public List<String> decode(String s) {
        List<String> res = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            int slash = s.indexOf('/', i);
            int len = Integer.valueOf(s.substring(i, slash));
            res.add(s.substring(slash + 1, slash + 1 + len));
            i = slash + len;
        }
        return res;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.decode(codec.encode(strs));
```

## Analysis

时间复杂度O(N).