[Leetcode] 408. Valid Word Abbreviation 解题报告

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.

思路：

一道练手题目，注意边界控制即可。

代码：

class Solution {public:    bool validWordAbbreviation(string word, string abbr) {        int index1 = 0, index2 = 0;        while(index1 < word.length() && index2 < abbr.length()){            if(isdigit(abbr[index2])) {                int value = abbr[index2++] - '0';                if (value == 0) {                    return false;                }                while(index2 < abbr.length() && isdigit(abbr[index2])) {                    value = 10 * value + abbr[index2++] - '0';                }                index1 += value;            }            else {                if(word[index1++] != abbr[index2++]) {                    return false;                }            }        }        if(index1 != word.length() || index2 != abbr.length()) {            return false;        }        return true;    }};