边工作边刷题：70天一遍leetcode: day 75

Group Shifted Strings

要点：开始就想到了string之间前后字符diff要相同。

• 思维混乱的地方：和某个string的diff之间是没有关系的。所以和单个string是否在那个点会出现z->a也没关系。
• 唯一tricky的地方是z->a的diff为-25，其和1是等价的。同理a->z diff为25，其和-1是等价的。显然差值是+/-26，两个值map到一个value的方法是(x+26)%26，+26是为了变到positive，%26是26循环。实际上可以用26 + val if val < 0 else val

错误点：

• 因为单个字符的范围是0-25，所以encoding要加delimiter
• python的get(key, default)得到的是临时object，而不会put internal value，要用collections.defaultdict()
• int() vs ord()

https://repl.it/Cive/2

# Given a string, we can "shift" each of its letter to its successive letter, for example: "abc" -> "bcd". We can keep "shifting" which forms the sequence:

# "abc" -> "bcd" -> ... -> "xyz"
# Given a list of strings which contains only lowercase alphabets, group all strings that belong to the same shifting sequence.

# For example, given: ["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"], 
# A solution is:

# [
#   ["abc","bcd","xyz"],
#   ["az","ba"],
#   ["acef"],
#   ["a","z"]
# ]
# Hide Company Tags Google Uber
# Hide Tags Hash Table String
# Hide Similar Problems (M) Group Anagrams

from collections import defaultdict
class Solution(object):
    def groupStrings(self, strings):
        """
        :type strings: List[str]
        :rtype: List[List[str]]
        """
        def getDiff(s):
            diff = []
            for pair in zip(s, s[1:]):
                u, v = pair[0], pair[1]
                # print u,v
                val = ord(v)-ord(u)
                diff.append(str(val+26 if val<0 else val))
            return '$'.join(diff)
            
        groups = defaultdict(list)
        single = []
        for s in strings:
            if len(s)==1:
                single.append(s)
            else:
                groups[getDiff(s)].append(s)
        ret = groups.values()
        if single:
            ret.append(single)
        return ret
   
sol = Solution()
assert sol.groupStrings(["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"])==[['az', 'ba'], ['acef'], ['abc', 'bcd', 'xyz'], ['a', 'z']]