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 :
Given s = "internationalization", abbr = "i12iz4n":
Solution：TwoPointers对比Check, 跳过Abbr
思路： i, j 分别遍历word 和 abbr，若元素相同，继续；若abbr开始出现数字num，则i跳过i += num，判断最后是否分别都到达串尾。
Time Complexity: O(N) Space Complexity: O(1)
Solution Code：
class Solution {
public boolean validWordAbbreviation(String word, String abbr) {
int i = 0, j = 0;
while (i < word.length() && j < abbr.length()) {
if (word.charAt(i) == abbr.charAt(j)) {
++i;++j;
continue;
}
if (abbr.charAt(j) <= '0' || abbr.charAt(j) > '9') {
return false;
}
int start = j;
while (j < abbr.length() && abbr.charAt(j) >= '0' && abbr.charAt(j) <= '9') {
++j;
}
int num = Integer.valueOf(abbr.substring(start, j));
i += num;
}
return i == word.length() && j == abbr.length();
}
}
