LeetCode - One Edit Distance

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

1. 两个字符串的长度之差大于1，那么直接返回False

2. 两个字符串的长度之差等于1，那么长的那个字符串去掉一个字符，剩下的应该和短的字符串相同

3. 两个字符串的长度之差等于0，那么两个字符串对应位置的字符只能有一处不同。

public class Solution {
    
    public static void main(String[] args){
        String s = "test1";
        String t = "test12";
        System.out.println(isOneEditDistance(s,t));
    }
    public static boolean isOneEditDistance(String s, String t) {
        int sl = s.length();
        int tl = t.length();
        if(Math.abs(sl - tl)>1){
            return false;
        }
        if(sl - tl == 1){
            return isDeleteOne(s, t);
        }
        if(tl - sl == 1){
            return isDeleteOne(t,s);
        }
        if(tl == sl){
            return isEditOne(s, t);
        }
        return false;   
    }
    
    public static boolean isDeleteOne(String sLong, String sShort){
        for(int i=0; i<sLong.length(); i++){
            String subStr = "";
            if(i == 0){
                subStr = sLong.substring(1);
            }
            else if(i == sLong.length()-1){
                subStr = sLong.substring(0,sLong.length()-1);
            }
            else{
                subStr = sLong.substring(0,i)+sLong.substring(i+1,sLong.length()-1);
            }
            if(subStr.equals(sShort)){
                return true;
            }
        }
        return false;
    }
    public static boolean isEditOne(String s, String t){
        boolean modified = false;
        for (int i=0; i<s.length(); i++){
            if(s.charAt(i) != t.charAt(i)){
                if(modified){
                    return false;
                }
                modified = true;
            }
        }
        return modified;
    }
}