408. Valid Word Abbreviation

No trick. Pretty straightforward solution. Note when return, we need to check if we've reached both ends of word and it abbreviation.

1:  class Solution {  
2:  public:  
3:    bool validWordAbbreviation(string word, string abbr) {  
4:      int i = 0, j = 0;  
5:      while (i < word.size() && j < abbr.size()) {  
6:        if (!isNum(abbr[j])) {  
7:          if (word[i++] != abbr[j++]) return false;  
8:        } else {  
9:          if (abbr[j] == '0') return false;  
10:          int l = 0;  
11:          while (j < abbr.size() && isNum(abbr[j])) {  
12:            l = l * 10 + abbr[j++] - '0';  
13:          }  
14:          i += l;  
15:        }  
16:      }  
17:      return i == word.size() && j == abbr.size();  
18:    }  
19:    bool isNum(char c) {  
20:      return c >= '0' && c <= '9';  
21:    }  
22:  };