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
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.--------------------
分三种情况:
1. 如果是数字,吃数字
2. 如果count == 0, 说明大家都指向字符。那就对比字符
3. 如果不是数字,j += count。 此时i, j都会指向字符或超出范围,两者对比会交给下一次的循环。
注意几个corner case:
1. 数字开头为0: a0b
2. 末尾是数字: ab1。 最后检测长度的时候加上count
class Solution { public boolean validWordAbbreviation(String word, String abbr) { int i = 0, j = 0; int count = 0; while (i < abbr.length() && j < word.length()) { char c = abbr.charAt(i); if (Character.isDigit(c)) { if (count == 0 && c <= '0') return false; count = Character.getNumericValue(c) + count * 10; i++; }else if (count == 0) { if (c != word.charAt(j)) return false; i++; j++; }else { j += count; count = 0; } } return i == abbr.length() && j + count == word.length(); } }
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