leetcode - 408. Valid Word Abbreviation

Description

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.

Solution

Sweep the abbreviation, if it's a number, continue. If it's a character, compare.

Time complexity: o ( n ) o(n) o(n)

Space complexity: o ( 1 ) o(1) o(1)

Code

python3 复制代码

class Solution:
    def validWordAbbreviation(self, word: str, abbr: str) -> bool:
        word_index = 0
        num = 0
        for each_c in abbr:
            if each_c.isdigit():
                if num == 0 and each_c == '0':
                    return False
                num = num * 10 + int(each_c)
            else:
                word_index += num
                num = 0
                if word_index >= len(word) or word[word_index] != each_c:
                    return False
                word_index += 1
        word_index += num
        return word_index == len(word)

Version2

python3 复制代码

class Solution:
    def validWordAbbreviation(self, word: str, abbr: str) -> bool:
        # i: index for word, j: index for abbr
        i, j = 0, 0
        num = 0
        while i < len(word) and j < len(abbr):
            if word[i] == abbr[j]:
                i += 1
                j += 1
            else:
                if not abbr[j].isdigit() or abbr[j] == '0':
                    return False
                num = 0
                while j < len(abbr) and '0' <= abbr[j] <= '9':
                    num = num * 10 + int(abbr[j])
                    j += 1
                i += num
        return i == len(word) and j == len(abbr)