有效单词缩写

题目

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.

总体来说题目的意思就是给定一个字符串,以下的形式是合法的缩写:

["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]

要注意仅有这些是合法的,其他都是非法的.

注意:所有的缩写和给定的字符都是小写的.

例1:

Given s = "internationalization", abbr = "i12iz4n":

Return true.

例2

Given s = "apple", abbr = "a2e":

Return false.

思路

我们观察到也就是字符还是字符的位置,中间的数字就是代表了经过的位数.

可以分成两个指针.一个指针指向word,一个指针指向abbr.如果是相同字符,则同时移动.如果abbr是数字,则将abbr的指针移动.然后使用截取得到数字,将word的指针直接移动.如果是不相同字符,则直接返回.最后只需要比较长度与原长度是否相等即可.

代码

class Solution {
   public boolean validWordAbbreviation(String word, String abbr) {
            int wordIndex = 0;
            int abbrIndex = 0;
            int wordLength = word.length();
            int abbrLength = abbr.length();
            //如果abbr的长度已经比word要长,则肯定为false
            if(abbrLength > wordLength){
                    return false;
            }
            while (abbrIndex < abbrLength && wordIndex < wordLength) {
                if (word.charAt(wordIndex) == abbr.charAt(abbrIndex)) {
                    wordIndex++;
                    abbrIndex++;
                } else if (Character.isDigit(abbr.charAt(abbrIndex))) {
                    int start = abbrIndex;
                    while (abbrIndex < abbr.length() && Character.isDigit(abbr.charAt(abbrIndex))) {
                        abbrIndex++;
                    }
                    int count = Integer.valueOf(abbr.substring(start, abbrIndex));
                    wordIndex += count;
                } else {
                    return false;
                }
            }
            return wordIndex == wordLength && abbrIndex == abbrLength;
        }