[Leetcode] 408. Valid Word Abbreviation 解题报告

最新推荐文章于 2020-04-01 23:12:37 发布

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本文介绍了一种用于判断字符串与其缩写是否匹配的算法。通过遍历字符串与缩写，利用数字跳过相应字符数的方式实现高效比对。适用于如'word'与其有效缩写如'1ord'、'wo2'等的验证。

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;
    }
};