LeetCode 161. One Edit Distance

原题链接在这里：https://leetcode.com/problems/one-edit-distance/

题目：

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:

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.

题解：

若是长度相差大于1, return false. 若是长度相差等于1, 遇到不同char时, 长的那个向后挪一位. 若是长度相等, 遇到不同char时同时向后挪一位.

出了loop还没有返回，就是到目前为止都相同，那么看长度差是不是等于1. 这里等于0表示完全相同也不可以.

Time Complexity: O(Math.min(len1, len2)).

Space: O(1).

AC Java:

 1 public class Solution {
 2     public boolean isOneEditDistance(String s, String t) {
 3         if(s == null || t == null){
 4             return false;
 5         }
 6         int len1 = s.length();
 7         int len2 = t.length();
 8         for(int i = 0; i < Math.min(len1, len2); i++){
 9             if(s.charAt(i) != t.charAt(i)){
10                 if(len1 == len2){
11                     return s.substring(i+1).equals(t.substring(i+1));
12                 }else if(len1 > len2){
13                     return s.substring(i+1).equals(t.substring(i));
14                 }else{
15                     return s.substring(i).equals(t.substring(i+1));
16                 }
17             }
18         }
19         return Math.abs(len1-len2) == 1;
20     }
21 }