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.
思路：算法就是：i， j兩個指針，j遇見數字，跳過，並且收集數字，然後i跳過這些數字，繼續走；注意的是：第一次不相等的時候，j不能爲0，後面j可以是0；
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;
}
// 不相等的第一個，不能爲0；
if(abbr.charAt(j) <= '0' || abbr.charAt(j) > '9') {
return false;
}
int start = j;
// 不相等的後面，可以是0；
while(j < abbr.length() && '0' <= abbr.charAt(j) && abbr.charAt(j) <= '9') {
j++;
}
int num = Integer.parseInt(abbr.substring(start, j));
i += num;
}
return i == word.length() && j == abbr.length();
}
}