Leetcode408. Valid Word Abbreviation
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题目链接:https://leetcode.com/contest/7/problems/valid-word-abbreviation/
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.
思路:
easy 要注意两点:1。连续的digit要组合为一个number 2。number不能有前导0,且number不为0
算法:
public boolean validWordAbbreviation(String word, String abbr)
int i = 0, j = 0, start = -1;
while (i < word.length() && j < abbr.length())
if (Character.isDigit(abbr.charAt(j)))
if (start == -1)
start = j;
if (abbr.charAt(j) - '0' == 0)
return false;
if (j == abbr.length() - 1)
int num = Integer.parseInt(abbr.substring(start, j + 1));
i += num;
j++;
else
if (start != -1)
int num = Integer.parseInt(abbr.substring(start, j));
i += num;
start = -1;
else
if (word.charAt(i) == abbr.charAt(j))
i++;
j++;
else
return false;
if (i == word.length() && j == abbr.length())
return true;
else
return false;
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408. Valid Word Abbreviation --字符串处理