Given two strings S and T, determine if they are both one edit distance apart.
Solution：
思路：
A = "1 2 3 4 5"
B = "1 2 3 4 5"
One Edit Check三种情况即可：
(1)delete one char in A，(2)delete one char in B, (3)replace one char in A
InsertA和deleteB效果相同
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) {
int len = Math.min(s.length(), t.length());
for(int c = 0; c < len; c++) {
if(s.charAt(c) != t.charAt(c)) {
// try if replacing works
if(s.length() == t.length()) return s.substring(c + 1).equals(t.substring(c + 1));
else {
// try if deleting works
if(s.length() > t.length()) return t.substring(c).equals(s.substring(c + 1)); //delete from s
else return s.substring(c).equals(t.substring(c + 1)); //delete from t
}
}
}
//All previous chars checked are the same, then check if the diff of lenth for the rest is one
return Math.abs(s.length() - t.length()) == 1;
}
}
