408. Valid Word Abbreviation

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.

Example 1:

Input: word = "internationalization", abbr = "i12iz4n"
Output: true
Explanation: The word "internationalization" can be abbreviated as "i12iz4n" ("i nternational iz atio n").

Example 2:

Input: word = "apple", abbr = "a2e"
Output: false
Explanation: The word "apple" cannot be abbreviated as "a2e".

Constraints:

• 1 <= word.length <= 20

• word consists of only lowercase English letters.

• 1 <= abbr.length <= 10

• abbr consists of lowercase English letters and digits.

• All the integers in abbr will fit in a 32-bit integer.

分析

遇到数字叠加，注意开头0的情况 用NUM==0 AND C == '0'判断

遇到字母， 更新POS比较

注意不要忘了最后的NUMBER!!!!

class Solution:
    def validWordAbbreviation(self, word: str, abbr: str) -> bool:
        # Input: word = "internationalization", abbr = "i12iz4n"
        pos = 0
        num = 0
        for i,c in enumerate(abbr):
            if c.isdigit():
                if num == 0 and c == '0':
                    return False
                num = num * 10 + int(c)
            else:
                pos += num
                num = 0
                if pos >= len(word) or word[pos] != c:
                    return False
                pos += 1
        return pos + num == len(word) #别忘了最后的NUMBER！！！！！！！