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408. Valid Word Abbreviation.java
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408. Valid Word Abbreviation.java
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1557989899
tags:String, Basic Implementation
tricky: find integer within a string
edge case: leading '0' should not be allow in such abbr.
```
/*
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.
*/
/*
abbr uses int to replace letters.
walk through abbr, if non-int letter from abbr all matches original word, and travese completes, it's a true.
*/
class Solution {
public boolean validWordAbbreviation(String word, String abbr) {
if (abbr == null || abbr.length() > word.length()) {
return false;
}
int index = 0; // index tracker on word
int i = 0; // index tracker for abbr
int m = word.length(), n = abbr.length();
while (i < n && index < m) {
if (word.charAt(index) == abbr.charAt(i)) {
index++;
i++;
continue;
}
// edge case: no integer, or: leading '0' cannot be used for abbr
if (!isInt(abbr.charAt(i)) || abbr.charAt(i) == '0') {
return false;
}
// obtain the int
StringBuffer sb = new StringBuffer();
while (i < n && isInt(abbr.charAt(i))) {
sb.append(abbr.charAt(i));
i++;
}
index += Integer.parseInt(sb.toString());
}
return index == m && i == n;
}
private boolean isInt(char c) {
return c >= '0' && c <= '9';
}
}
```