[leetcode]161. One Edit Distance编辑步数为一

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.

题目

给定两个字符串，判断其编辑步数是否为1

思路

此题可算是[leetcode]72. Edit Distance 最少编辑步数的一个拆分简化版本

代码

 class Solution {
      public boolean isOneEditDistance(String s, String t) {
         int m = s.length(), n = t.length();
         if(m == n) return isOneModified(s, t);
         if(m - n == 1) return isOneDeleted(s, t);
         if(n - m == 1) return isOneDeleted(t, s);
         // 长度差距大于2直接返回false
         return false;
     }
 
     private boolean isOneModified(String s, String t){
         boolean modified = false;
         // 看是否只修改了一个字符
         for(int i = 0; i < s.length(); i++){
             if(s.charAt(i) != t.charAt(i)){
                 if(modified) return false;
                 modified = true;
             }
         }
         return modified;
     }
 
     public boolean isOneDeleted(String longer, String shorter){
         // 找到第一组不一样的字符，看后面是否一样
         for(int i = 0; i < shorter.length(); i++){
             if(longer.charAt(i) != shorter.charAt(i)){
                 return longer.substring(i + 1).equals(shorter.substring(i));
             }
         }
         return true;
     }
 }