### LeetCode  • ㊗️
• 大家
• offer
• 多多！

## Problem

Given two strings s and t, determine if they are both one edit distance apart.

Note:

There are 3 possibilities to satisfy one edit distance apart:

Insert a character into s to get t Delete a character from s to get t Replace a character of s to get t Example 1:

Input: s = "ab", t = "acb"
Output: true
Explanation: We can insert 'c' into s to get t.


Example 2:

Input: s = "cab", t = "ad"
Output: false
Explanation: We cannot get t from s by only one step.


Example 3:

Input: s = "1203", t = "1213"
Output: true
Explanation: We can replace '0' with '1' to get t.


## Code

class Solution {
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()){
return s.substring(i + 1).equals(t.substring(i + 1));
} else if(s.length() < t.length()){
return s.substring(i).equals(t.substring(i + 1));
} else if(s.length() > t.length()){
return s.substring(i + 1).equals(t.substring(i));
}
}
}

return Math.abs(s.length() - t.length()) == 1;
}
}