[LeetCode] Valid Word Abbreviation 验证单词缩写

 

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.

 

这道题让我们验证单词缩写,关于单词缩写LeetCode上还有两道相类似的题目Unique Word AbbreviationGeneralized Abbreviation。这道题给了我们一个单词和一个缩写形式,让我们验证这个缩写形式是否是正确的,由于题目中限定了单词中只有小写字母和数字,所以我们只要对这两种情况分别处理即可。我们使用双指针分别指向两个单词的开头,循环的条件是两个指针都没有到各自的末尾,如果指向缩写单词的指针指的是一个数字的话,如果当前数字是0,返回false,因为数字不能以0开头,然后我们要把该数字整体取出来,所以我们用一个while循环将数字整体取出来,然后指向原单词的指针也要对应的向后移动这么多位数。如果指向缩写单词的指针指的是一个字母的话,那么我们只要比两个指针指向的字母是否相同,不同则返回false,相同则两个指针均向后移动一位,参见代码如下:

 

解法一:

class Solution {
public:
    bool validWordAbbreviation(string word, string abbr) {
        int i = 0, j = 0, m = word.size(), n = abbr.size();
        while (i < m && j < n) {
            if (abbr[j] >= '0' && abbr[j] <= '9') {
                if (abbr[j] == '0') return false;
                int val = 0;
                while (j < n && abbr[j] >= '0' && abbr[j] <= '9') {
                    val = val * 10 + abbr[j++] - '0';
                }
                i += val;
            } else {
                if (word[i++] != abbr[j++]) return false;
            }
        }
        return i == m && j == n;
    }
};

 

下面这种方法和上面的方法稍有不同,这里是用了一个for循环来遍历缩写单词的所有字符,然后用一个指针p来指向与其对应的原单词的位置,然后cnt表示当前读取查出来的数字,如果读取的是数字,我们先排除首位是0的情况,然后cnt做累加;如果读取的是字母,那么指针p向后移动cnt位,如果p到超过范围了,或者p指向的字符和当前遍历到的缩写单词的字符不相等,则返回false,反之则给cnt置零继续循环,参见代码如下:

 

解法二:

class Solution {
public:
    bool validWordAbbreviation(string word, string abbr) {
        int m = word.size(), n = abbr.size(), p = 0, cnt = 0;
        for (int i = 0; i < abbr.size(); ++i) {
            if (abbr[i] >= '0' && abbr[i] <= '9') {
                if (cnt == 0 && abbr[i] == '0') return false;
                cnt = 10 * cnt + abbr[i] - '0';
            } else {
                p += cnt;
                if (p >= m || word[p++] != abbr[i]) return false;
                cnt = 0;
            }
        }
        return p + cnt == m;
    }
};

 

类似题目:

Unique Word Abbreviation

Generalized Abbreviation

 

参考资料:

https://discuss.leetcode.com/topic/61404/concise-c-solution

https://discuss.leetcode.com/topic/61430/java-2-pointers-15-lines

https://discuss.leetcode.com/topic/61353/simple-regex-one-liner-java-python

 

    原文作者:Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/5930369.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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