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:
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| ["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]
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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:
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| Given s = "internationalization", abbr = "i12iz4n":
Return true.
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Example 2:
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| Given s = "apple", abbr = "a2e":
Return false.
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| public class Solution {
public boolean validWordAbbreviation(String word, String abbr) {
Map<Integer, Character> map = new HashMap<>();
int length = parse(abbr, map);
if (word.length() != length) {
return false;
}
for (Map.Entry<Integer, Character> entry : map.entrySet()) {
if (!Objects.equals(word.charAt(entry.getKey()), entry.getValue())) {
return false;
}
}
return true;
}
private int parse(String abbr, Map<Integer, Character> map) {
int length = 0;
int x = 0;
for (int i = 0; i < abbr.length(); i++) {
char c = abbr.charAt(i);
if (c >= 'a' && c <= 'z') {
length += x + 1;
map.put(length-1, c);
x = 0;
} else {
if(x == 0 && c == '0') return -1;
x = 10 * x + c - '0';
}
}
if (x > 0) {
length += x;
}
return length;
}
}
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