My code:

public class Codec {

    // Encodes a list of strings to a single string.
    public String encode(List<String> strs) {
        if (strs == null || strs.size() == 0) {
            return null;
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < strs.size(); i++) {
            sb.append(strs.get(i).length() + "/" + strs.get(i));
        }
        return sb.toString();
    }

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

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

reference:
https://discuss.leetcode.com/topic/22848/ac-java-solution

看的答案，的确巧妙。
前面放长度，后面放string，中间用 ‘/’ 分隔开
本来觉得任何分隔符都会出现，那会造成干扰。其实不会。
比如encode “/”
=> “1//”
decode “1//”
=> “/”

然后，string和前面的范围不能交换。必须先有范围，才能找到后面的string。

Anyway, Good luck, Richardo! — 09/20/2016