leetcode408 - Valid Word Abbreviation - easy

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.

 

 

两根同向指针。

一根指着abbr一根指着word，ptrA发现数字的时候要停下来，取数字，帮ptrW也跳这么远，然后接着比。最后确认一下两根指针是不是都跑到各自的最后了即可。

 

细节：

1.API。isLetter(()方法的正确写法是Character.isLetter(abbr.charAt(pa))而不是abbr.charAt(pa).isLetter()，char没有方法，只有Character有static方法

2.abbr里的0只有101这种数字里才算有效，以0开头不行，比如"a" "01"这种case应输出false，因为题意里说的word只有w1rd这几种形式，而没有w01rd的形式，可以说算是给的比较明确了。具体当然和面试官沟通。 

 

我的实现：

class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        int pw = 0;
        int pa = 0;
        while (pw < word.length() && pa < abbr.length()) {
            // 注意API！不是abbr.charAt(pa).isLetter()，char没有方法，只有Character有static方法
            if (Character.isLetter(abbr.charAt(pa))) {
                if (pw >= word.length() || word.charAt(pw) != abbr.charAt(pa)) {
                    return false;
                }
                pw++;
                pa++;
            } else if (abbr.charAt(pa) == '0') {
                // 注意0不能算，"a" "01"这种case应输出false，具体当然和面试官沟通。
                return false;
            } else {
                int number = 0;
                while (pa < abbr.length() && Character.isDigit(abbr.charAt(pa))) {
                    number = number * 10 + abbr.charAt(pa) - '0';
                    pa++;
                }
                pw += number;
            }
        }
        return pw == word.length() && pa == abbr.length();
    }
}

九章实现

public class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        int i = 0, j = 0;
        while (i < word.length() && j < abbr.length()) {
            if (word.charAt(i) == abbr.charAt(j)) {
                i++;
                j++;
            } else if ((abbr.charAt(j) > '0') && (abbr.charAt(j) <= '9')) {     //notice that 0 cannot be included
                int start = j;
                while (j < abbr.length() && Character.isDigit(abbr.charAt(j))) {
                    j++;
                }
                i += Integer.valueOf(abbr.substring(start, j));
            } else {
                return false;
            }
        }
        return (i == word.length()) && (j == abbr.length());
    }
}