161.One Edit Distance

1- https://discuss.leetcode.com/topic/22435/easy-understood-java-solution

class Solution {
    public boolean isOneEditDistance(String s, String t) {
        int lenS = s.length();
        int lenT = t.length();
        if(Math.abs(lenS - lenT) > 1)
            return false;
        if(lenS == lenT)
            return isOneModify(s, t);
        if(lenS > lenT)
            return isOneDel(s, t);
        else
            return isOneDel(t, s);
    }

    public boolean isOneModify(String s, String t){
        int diff = 0;
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i) != t.charAt(i))
                diff++;
        }
        return diff == 1;
    }
    public boolean isOneDel(String l, String s){
        for(int i = 0, j = 0; i < l.length() && j < s.length(); i++, j++){
            if(s.charAt(i) != l.charAt(i))
                return l.substring(i + 1).equals(s.substring(j));
        }
        return true;
    }
}

2

/*
 * There're 3 possibilities to satisfy one edit distance apart: 
 * 
 * 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 boolean isOneEditDistance(String s, String t) {
    for (int i = 0; i < Math.min(s.length(), t.length()); i++) { 
        if (s.charAt(i) != t.charAt(i)) {
            if (s.length() == t.length()) // s has the same length as t, so the only possibility is replacing one char in s and t
                return s.substring(i + 1).equals(t.substring(i + 1));
            else if (s.length() < t.length()) // t is longer than s, so the only possibility is deleting one char from t
                return s.substring(i).equals(t.substring(i + 1));                
            else // s is longer than t, so the only possibility is deleting one char from s
                return t.substring(i).equals(s.substring(i + 1));
        }
    }       
    //All previous chars are the same, the only possibility is deleting the end char in the longer one of s and t 
    return Math.abs(s.length() - t.length()) == 1;        
}