边工作边刷题：70天一遍leetcode: day 88
Unique Word Abbreviation
要点：
- 简单题，主要是理解题意。no other words have the same abbreviation as the given word意思就是如果没有同样的abbrev或者有但是只由dict中相同的word产生。
- 可以用True/False表示unique，同时用set来保存哪个word产生的abbrev
# An abbreviation of a word follows the form <first letter><number><last letter>. Below are some examples of word abbreviations:
# a) it --> it (no abbreviation)
# 1
# b) d|o|g --> d1g
# 1 1 1
# 1---5----0----5--8
# c) i|nternationalizatio|n --> i18n
# 1
# 1---5----0
# d) l|ocalizatio|n --> l10n
# Assume you have a dictionary and given a word, find whether its abbreviation is unique in the dictionary. A word's abbreviation is unique if no other word from the dictionary has the same abbreviation.
# Example:
# Given dictionary = [ "deer", "door", "cake", "card" ]
# isUnique("dear") -> false
# isUnique("cart") -> true
# isUnique("cane") -> false
# isUnique("make") -> true
class ValidWordAbbr(object):
def __init__(self, dictionary):
"""
initialize your data structure here.
:type dictionary: List[str]
"""
self.countmap = {}
self.words = set(dictionary)
for w in self.words:
abw = self.getAbbre(w)
self.countmap[abw]=True if abw not in self.countmap else False
def isUnique(self, word):
"""
check if a word is unique.
:type word: str
:rtype: bool
"""
w = self.getAbbre(word)
return w not in self.countmap or (self.countmap[w] and word in self.words)
def getAbbre(self, word):
n = len(word)
if n<=2: return word
return word[0]+str(n-2)+word[n-1]
# Your ValidWordAbbr object will be instantiated and called as such:
# vwa = ValidWordAbbr(dictionary)
# vwa.isUnique("word")
# vwa.isUnique("anotherWord")
