Given two strings S and T, determine if they are both one edit distance apart.

一刷
題解：只會有三種情況

1. Replace 1 char:
   s: a B c
   t: a D c
2. Delete 1 char from s:
   s: a D b c
   t: a b c
3. Delete 1 char from t
   s: a b c
   t: a D b c

public class Solution {
    public boolean isOneEditDistance(String s, String t) {
        int len = Math.min(s.length(), t.length());
        for(int i=0; i<len; i++){
            if(s.charAt(i) != t.charAt(i)){
                if(s.length() == t.length()){
                    return s.substring(i+1).equals(t.substring(i+1));
                }
                if(s.length()<t.length()){//delete one char in s
                    return s.substring(i).equals(t.substring(i+1));
                }
                else{
                    return t.substring(i).equals(s.substring(i+1));
                }
            }
        }
        return Math.abs(s.length() - t.length())==1;
    }
}

二刷
思路同上

public class Solution {
    public boolean isOneEditDistance(String s, String t) {
        int len = Math.min(s.length(), t.length());
        for(int i=0; i<len; i++){
            if(s.charAt(i)!=t.charAt(i)){
                if(s.length() == t.length()) return s.substring(i+1).equals(t.substring(i+1));
                else if(s.length() > t.length()) return s.substring(i+1).equals(t.substring(i));
                else return s.substring(i).equals(t.substring(i+1));
            }
        }
        
        return Math.abs(s.length()-t.length())==1;
    }
}```