题目描述:
LeetCode 408. Valid Word Abbreviation
Given a non-empty string s
and an abbreviation abbr
, return whether the string matches with the given abbreviation.
A string such as "word"
contains only the following valid abbreviations:
["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]
Notice that only the above abbreviations are valid abbreviations of the string "word"
. Any other string is not a valid abbreviation of "word"
.
Note:
Assume s
contains only lowercase letters and abbr
contains only lowercase letters and digits.
Example 1:
Given s = "internationalization", abbr = "i12iz4n": Return true.
Example 2:
Given s = "apple", abbr = "a2e": Return false.
题目大意:
给定一个非空字符串s以及一个单词缩写abbr,返回字符串是否与单词缩写相匹配。
一个诸如"word"的字符串只包含如下有效的缩写:
["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]
请注意只有上面的缩写是"word"的有效缩写。其他任何字符串都不是"word"的有效缩写。其他任何字符串都不是
注意:
假设s只包含小写字母,abbr只包含小写字母和数字。
解题思路:
模拟题,遍历word和abbr即可。
Python代码:
class Solution(object):
def validWordAbbreviation(self, word, abbr):
"""
:type word: str
:type abbr: str
:rtype: bool
"""
size = len(word)
cnt = loc = 0
for w in abbr:
if w.isdigit():
if w == '0' and cnt == 0:
return False
cnt = cnt * 10 + int(w)
else:
loc += cnt
cnt = 0
if loc >= size or word[loc] != w:
return False
loc += 1
return loc + cnt == size
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