161. One Edit Distance

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

Solution：String

思路：
Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

class Solution {
    /*
     * 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;        
    }
}