Description
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.
Solution
Two-pointer, O(n), S(1)
簡單題，但是testcase有點(diǎn)坑，并沒有保證abbr一定是合法的恢氯，例如：
"a"
"0a"
需要return false...
class Solution {
public boolean validWordAbbreviation(String word, String abbr) {
if (word == null || abbr == null) {
return false;
}
int i = 0;
int j = 0;
int m = word.length();
int n = abbr.length();
while (j < n && i < m) {
char c = abbr.charAt(j);
if (Character.isLetter(c)) {
if (word.charAt(i++) != abbr.charAt(j++)) {
return false;
}
} else {
if (c == '0') { // leading 0 is invalid
return false;
}
int num = 0;
while (j < n && Character.isDigit(abbr.charAt(j))) {
num = 10 * num + abbr.charAt(j++) - '0';
}
i += num;
}
}
return j == n && i == m;
}
}
