感冒之后
睡了2天觉
现在痊愈了
重启刷题进程。。
Google的题，E难度。。
比较的方法很多，应该是为后面的题铺垫的。
题不难，做对不容易，edge cases很多，修修改改好多次，写完发现是一坨。
奉劝大家尽量多想edge cases，想不出来了再提交，别像我似的赶着投胎就提交了。
想完了先自己试试下面这些test cases...
"internationalization"
"i12iz4n"
"word"
"1or1"
"apple"
"a2e"
"a"
"2"
"b"
"1"
"a"
"01"
"hi"
"hi1"
"hi"
"2i"
public class Solution
{
public boolean validWordAbbreviation(String word, String abbr)
{
if(word.length() == 0 && abbr.length() == 0) return true;
if(word.length() == 0 || abbr.length() == 0) return false;
int m = 0; int n = 0;
int digit = 0;
while(m < word.length() && n < abbr.length())
{
if(abbr.charAt(n) >= '0' && abbr.charAt(n) <= '9')
{
if(abbr.charAt(n) == '0' && digit == 0) return false; // digit starts with 0
digit = 10*digit + abbr.charAt(n) - '0';
n++;
if(n == abbr.length()) // if abbr ends, word has to end at the same time
{
return m + digit == word.length();
}
}
else
{
m += digit;
digit = 0;
if(m == word.length())
{
if(n == abbr.length()) return true; // both end
else return false; // word ends, abbr not
}
if(m > word.length()) return false; // abbr is longer than word
if(word.charAt(m) != abbr.charAt(n)) return false; // match fails
else // go on...
{
m++;
n++;
}
}
}
//not even same length
return m >= word.length() && n >= abbr.length();
}
}