how to efficiently check if the Levenshtein edit distance between two string is 1 [closed]

问题

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.

Closed 7 years ago.

please note that it doesn't require to really calculate Levenshtein edit distance. just check it's 1 or not.

The signature of the method may look like this:

bool Is1EditDistance(string s1, string s2). 

for example: 1. "abc" and "ab" return true 2. "abc" and "aebc" return true 3. "abc" and "a" return false.

I've tried recursive approve, but it it not efficient.

---

update: got answer from a friend:

        for (int i = 0; i < s1.Length && i < s2.Length; i++)
        {
            if (s1[i] != s2[i])
            {
                return s1.Substring(i + 1) == s2.Substring(i + 1)   //case of change
                    || s1.Substring(i + 1) == s2.Substring(i)       //case of s1 has extra
                    || s1.Substring(i) == s2.Substring(i + 1);      //case of s2 has extra
            }
        }
        return Math.Abs(s1.Length - s2.Length) == 1;

回答1:

If you only care if the distance is exactly 1 or not, you can do something like this:

• If the difference of the strings' lengths is not 0 or 1, return false.
• If both strings have length n, loop i = 0..n checking s1[i] == s2[i] for all i except one.
• If the strings have length n and n+1, let i be the smallest index where s1[i] != s2[i], then loop j=i..n checking s1[j] == s2[j+1] for all j.

来源：https://stackoverflow.com/questions/12632825/how-to-efficiently-check-if-the-levenshtein-edit-distance-between-two-string-is