题目
A string can be abbreviated by replacing any number of non-adjacent, non-empty substrings with their lengths. The lengths should not have leading zeros.
For example, a string such as “substitution” could be abbreviated as (but not limited to):
“s10n” (“s ubstitutio n”) “sub4u4” (“sub stit u tion”) “12” (“substitution”) “su3i1u2on” (“su bst i t u ti on”) “substitution” (no substrings replaced) The following are not valid abbreviations:
“s55n” (“s ubsti tutio n”, the replaced substrings are adjacent) “s010n” (has leading zeros) “s0ubstitution” (replaces an empty substring) Given a string word and an abbreviation abbr, return whether the string matches the given abbreviation.
A substring is a contiguous non-empty sequence of characters within a string.
解题思路
两个指针指向字符串和缩写，同时向右移动，如果缩写是数字，就记录下来记录当前数字，直到不是数字为止，遇到字母，指向单词的指针向右移动数字位置，判断不能越界且对应的字母相等，最后处理缩写尾部是数字的情况，并且两个指针都到走到最后
代码
class Solution:
def validWordAbbreviation(self, word: str, abbr: str) -> bool:
m, n = len(word), len(abbr)
i = j = cur = 0
while i < m and j < n:
if abbr[j].isdigit():
if abbr[j] == "0" and cur == 0:
return False
cur = cur * 10 + int(abbr[j])
else:
i += cur
cur = 0
if i >= m or word[i] != abbr[j]:
return False
i += 1
j += 1
return i + cur == m and j == n
