[LeetCode]161. Edit distance of 1

subject

 Given two strings  s  and  t, Judge whether their editing distance is  1.

 Be careful ：
 The editing distance is equal to  1  There are three possible situations ：
 Go to  s  Insert a character in to get  t
 from  s  Delete a character to get  t
 stay  s  Replace a character in to get  t

 Examples 

 Example  1：

 Input : s = "ab", t = "acb"
 Output : true
 explain :  Can be  'c'  Insert string  s  To get  t.

 Example  2:

 Input : s = "cab", t = "ad"
 Output : false
 explain :  Unable to get  1  Step operation  s  Turn into  t.

 Example  3:

 Input : s = "1203", t = "1213"
 Output : true
 explain :  You can string  s  Medium  '0'  Replace with  '1'  To get  t.

Method 1： Compare strings

public static void main(String[] args) {
    
    _1st handler = new _1st();
    String s = "ab", t = "acb";
    Assert.assertTrue(handler.isOneEditDistance(s, t));
    s = "cab";
    t = "ad";
    Assert.assertFalse(handler.isOneEditDistance(s, t));
    s = "1203";
    t = "1213";
    Assert.assertTrue(handler.isOneEditDistance(s, t));


}


public boolean isOneEditDistance(String s, String t) {
    
    int sn = s.length(), tn = t.length();
    // maintain s The length of is less than t
    if (sn > tn) {
    
        return isOneEditDistance(t, s);
    }
    if (tn - sn > 1) return false;// Greater than 1 when , return false
    for (int i = 0; i < sn; i++) {
    
        if (s.charAt(i) != t.charAt(i)) {
    
            //s And t The same length , Compare the following 
            if (sn == tn) {
    
                return s.substring(i + 1).equals(t.substring(i + 1));
            } else {
    //s And t The length is different  s The characters of are short 
                return s.substring(i).equals(t.substring(i + 1));
            }
        }
    }
    return sn + 1 == tn;
}