标签：continue case style tco substr empty lower app letters
原题链接在这里：https://leetcode.com/problems/valid-word-abbreviation/#/description
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.
题解：
双指针，当j指向数字时, i就跳过相应数字. 最后看i, j是否同时到尾部.
Time Complexity: O(n), n是word, abbr中较长的length.
AC Java:
1 public class Solution { 2 public boolean validWordAbbreviation(String word, String abbr) { 3 if(word == null && abbr == null){ 4 return true; 5 } 6 7 if(word == null || abbr == null){ 8 return false; 9 } 10 11 int i = 0; 12 int j = 0; 13 while(i<word.length() && j<abbr.length()){ 14 if(word.charAt(i) == abbr.charAt(j)){ 15 i++; 16 j++; 17 continue; 18 } 19 if(abbr.charAt(j)<=‘0‘ || abbr.charAt(j)>‘9‘){ 20 return false; 21 } 22 int s = j; 23 while(j<abbr.length() && abbr.charAt(j)>=‘0‘ && abbr.charAt(j)<=‘9‘){ 24 j++; 25 } 26 int jump = Integer.valueOf(abbr.substring(s, j)); 27 i += jump; 28 } 29 return i == word.length() && j == abbr.length(); 30 } 31 }
LeetCode Valid Word Abbreviation
标签：continue case style tco substr empty lower app letters
原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/6691768.html
