題目：

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.

鏈接：https://leetcode.com/problems/valid-word-abbreviation/#/description

3/22/2017

注意：

1. 第8行判斷是否有leading的0，我認為這個屬于invalid input。不過既然是google，需要仔細檢查所有可能。

2. 遇到char時，index如何加，比較之后是否加。其實，在比較之后，之前inInteger的狀態已經無所謂了，所以都需要+1

1 public class Solution { 2 public boolean validWordAbbreviation(String word, String abbr) { 3 boolean inInteger = false; 4 int number = 0; 5 int index = 0; 6 for (int i = 0; i < abbr.length(); i++) { 7 if (Character.isDigit(abbr.charAt(i))) { 8 if (!inInteger && abbr.charAt(i) - '0' == 0) return false; 9 number = number * 10 + abbr.charAt(i) - '0'; 10 inInteger = true; 11 } else { 12 if (inInteger) { 13 index += number; 14 inInteger = false; 15 number = 0; 16 } 17 if (index >= word.length()) return false; 18 if (word.charAt(index) != abbr.charAt(i)) return false; 19 index += 1; 20 } 21 } 22 if (inInteger) { 23 index += number; 24 } 25 if (index != word.length()) return false; 26 return true; 27 } 28 }

別人的思路：

1. 每次先比較2者是否相同，若不同查abbr是否是<=0或者>9（注意這里就排除了leading 0），再記錄abbr index遍歷abbr直到abbr的值不為數字，同時word加上中間的間隔。下次比較可以有結論。

2. 有個外國老哥，總是有很巧妙的方法：

1 public boolean validWordAbbreviation(String word, String abbr) { 2 return word.matches(abbr.replaceAll("[1-9]\\d*", ".{$0}")); 3 }

其他討論：https://discuss.leetcode.com/category/535/valid-word-abbreviation

轉載于:https://www.cnblogs.com/panini/p/6609229.html

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