边工作边刷题：70天一遍leetcode: day 81

Encode and Decode Strings

要点：题的特点：不是压缩，而是encode为字节流。所以需要找delimiter来分割每个word，但是delimiter可能是字符本身，所以可以用数字记录每个word的长度。这样一般比delimiter简单并且节省空间。接着可能会想到如果前后单词有数字怎么办？所以需要一个特殊字符来分割数字和词本身，而这个字符要在数字之后：不用担心前面单词以数字结尾，前一个单词通过长度信息已经成功decode。

• ''.join('%d:' % len(s)+s for s in strs)意思：(‘%d:’% len(s) + s) 这个对strs做list comprehension产生新list，所以每个都有len(s):在开头，new style是’’.join(‘{}:{}’.format(len(s), s) for s in strs

错误点：

• 要用find(‘:’, i)，也就是有boundary的方法来得到下一个:的位置
• 这题证明list.append => join 比str+=快不少
• 实际上，多string/int混合最好用formatter

https://repl.it/Cgvz/1

# 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.
# Hide Company Tags Google
# Hide Tags String
# Hide Similar Problems (E) Count and Say (H) Serialize and Deserialize Binary Tree


class Codec:

    def encode(self, strs):
        """Encodes a list of strings to a single string.
        
        :type strs: List[str]
        :rtype: str
        """
        return ''.join("%d:" % len(s) + s for s in strs)
        
    def decode(self, s):
        """Decodes a single string to a list of strings.
        
        :type s: str
        :rtype: List[str]
        """
        i = 0
        ret = []
        while i<len(s):
        	j = s.find(':', i) # error 1: should have i as find's left boundary, inclusive
        	i = j+int(s[i:j])+1
        	ret.append(s[j+1:i])
        
        return ret
        
sol = Codec()
assert sol.decode(sol.encode(['3e5d', 'd:4rf']))==['3e5d', 'd:4rf']