LintCode 640 - Edit Distance II

http://lintcode.com/en/problem/edit-distance-ii/

Given two strings S and T, determine if they are both one edit distance apart.

Example

Given s = "aDb", t = "adb"
return true

http://blog.csdn.net/sunday0904/article/details/71154703

首先判断两个字符串的长度，是否符合一个编辑距离。然后判断是要用替换还是插入。

1.     public boolean isOneEditDistance(String s, String t) {  
2.         int slen = s.length();  
3.         int tlen = t.length();  
4.         int temp = slen - tlen;  
5.         if(s.equals(t)) return false;  
6.         if(temp == 0){  
7.             boolean flag = true;  
8.             for(int i = 0; i< slen ;i++){  
9.                 if(s.charAt(i) != t.charAt(i)){  
10.                     if(flag){  
11.                         flag = false;  
12.                     }else{  
13.                         return false;  
14.                     }  
15.                 }  
16.             }  
17.             return true;  
18.         }else if(temp == 1){  
19.             if(s.substring(0 , tlen).equals(t)){  
20.                 return true;  
21.             }  
22.             boolean flag = true;  
23.             for(int i = 0;i<slen;i++){  
24.                 if(flag){  
25.                     if(s.charAt(i) != t.charAt(i)){  
26.                         flag = false;  
27.                     }      
28.                 }else{  
29.                     if(s.charAt(i) != t.charAt(i - 1)){  
30.                         return false;  
31.                     }  
32.                 }  
33.                   
34.             }  
35.             return true;  
36.         }else if(temp == -1){  
37.             if(t.substring(0 , slen).equals(s)){  
38.                 return true;  
39.             }  
40.             boolean flag = true;  
41.             for(int i = 0;i<tlen;i++){  
42.                 if(flag){  
43.                     if(t.charAt(i) != s.charAt(i)){  
44.                         flag = false;  
45.                     }      
46.                 }else{  
47.                     if(t.charAt(i) != s.charAt(i - 1)){  
48.                         return false;  
49.                     }  
50.                 }  
51.                   
52.             }  
53.             return true;  
54.         }else{  
55.             return false;  
56.         }  
57.     }