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

Note: 

There are 3 possiblities to satisify one edit distance apart:

1. Insert a character into s to get t
2. Delete a character from s to get t
3. 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.
 

題目

給定兩個字串，判斷其編輯步數是否為1

思路

此題可算是[leetcode]72. Edit Distance 最少編輯步數的一個拆分簡化版本

程式碼

 1 class Solution {
 2      public boolean isOneEditDistance(String s, String t) {
 3         int m = s.length(), n = t.length();
 4         if(m == n) return isOneModified(s, t);
 5         if
 
(m - n == 1) return isOneDeleted(s, t);
 6         if(n - m == 1) return isOneDeleted(t, s);
 7         // 長度差距大於2直接返回false
 8         return false;
 9     }
10     
11     private boolean isOneModified(String s, String t){
12         boolean modified = false;
13         // 看是否只修改了一個字元
14         for(int i = 0; i < s.length(); i++){
15             if(s.charAt(i) != t.charAt(i)){
16                 if(modified) return false;
17                 modified = true;
18             }
19         }
20         return modified;
21     }
22     
23     public boolean isOneDeleted(String longer, String shorter){
24         // 找到第一組不一樣的字元，看後面是否一樣
25         for(int i = 0; i < shorter.length(); i++){
26             if(longer.charAt(i) != shorter.charAt(i)){
27                 return longer.substring(i + 1).equals(shorter.substring(i));
28             }
29         }
30         return true;
31     }
32 }