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++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.
java
class Solution {
public boolean validWordAbbreviation(String word, String abbr) {
int i = 0;
int j = 0;
while(i<word.length() && j<abbr.length()){
char a = abbr.charAt(j);
if(Character.isDigit(a)){
if(a == '0'){
return false;
}
int number = a - '0'; //这里的number每次都要重新生成一下
while(j+1 < abbr.length() && Character.isDigit(abbr.charAt(j+1))){
number = number*10 + (abbr.charAt(j+1) - '0');
j++;
}
i += number;
j++;
}else{
if(abbr.charAt(j) != word.charAt(i)){
return false;
}else{
i++;
j++;
}
}
}
return i == word.length() && j == abbr.length(); //最后不是无脑返回true,要make sure所有的指针都走到了最后
}
}