[LeetCode] 408. Valid Word Abbreviation_Easy
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.
思路就是依次将abbr的数字弄出来, 然后再word上面相加, 否则就比较char, 最后出来loop之后应该是都指到各自的length上面.
Code
class Solution:
def validWordAbbreviation(self, word, abbr):
nums, lw, la, pw, pa = "", len(word), len(abbr),0,0
while pw < lw and pa < la:
num = ""
while abbr[pa] in nums:
num += abbr[pa]
if num[0] == '': return False
pa += 1
if pa == la:
return len(word[pw:]) == int(num)
if not num:
if word[pw] != abbr[pa]:
return False
else:
pw += 1
pa += 1
else:
pw += int(num)
return pw == lw and pa == la
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