Valid Word Abbreviation

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.

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这道题思路比较简单，但是后来处理边界条件上花了时间比较久。

思路是使用两个指针，分别从word, abbr的头开始走起，index2碰到数字的话，首先判断开头是否为0（这一点我挺费解的，为啥不能为0，后来测试用例告诉我这样不可以）。然后让index1 走相应的距离，然后再判断对应index1 index2的位置的值是否一样。不一样的话return false.

代码：

public boolean validWordAbbreviation(String word, String abbr) {        if(word == null || word.length() == 0 || abbr == null || abbr.length() == 0) return false;        int index1 = 0;        int index2 = 0;        int offset = 0;        while(index1<word.length() && index2 < abbr.length()){            char char2 = abbr.charAt(index2);            if(char2>='0' && char2 <='9'){                if(char2 == '0') return false;                while(char2>='0' && char2 <='9'){                    offset = offset * 10 + char2 - '0';                    index2++;                    if(index2 == abbr.length()) break;                    char2 = abbr.charAt(index2);                }            }            //System.out.println("offset"+ offset);            index1 += offset;            offset = 0;            if(index1 >= word.length()) break;            char char1 = word.charAt(index1);            if(char1 != char2){                return false;            }            index1++;            index2++;        }        return index1 == word.length() && index2 == abbr.length();    }