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6 | 6 | * Given two strings s and t, determine if they are both one edit distance apart. |
7 | 7 | * |
8 | 8 | * Note: |
9 | | - * |
10 | 9 | * There are 3 possiblities to satisify one edit distance apart: |
11 | | - * |
12 | | - * Insert a character into s to get t |
13 | | - * Delete a character from s to get t |
14 | | - * Replace a character of s to get t |
| 10 | + * Insert a character into s to get t |
| 11 | + * Delete a character from s to get t |
| 12 | + * Replace a character of s to get t |
15 | 13 | * |
16 | 14 | * Example 1: |
17 | 15 | * Input: s = "ab", t = "acb" |
|
29 | 27 | * Explanation: We can replace '0' with '1' to get t. |
30 | 28 | */ |
31 | 29 | public class _161 { |
32 | | - public static class Solution1 { |
33 | | - public boolean isOneEditDistance(String s, String t) { |
34 | | - char[] schar = s.toCharArray(); |
35 | | - char[] tchar = t.toCharArray(); |
| 30 | + public static class Solution1 { |
| 31 | + public boolean isOneEditDistance(String s, String t) { |
| 32 | + char[] schar = s.toCharArray(); |
| 33 | + char[] tchar = t.toCharArray(); |
36 | 34 |
|
37 | | - if (Math.abs(s.length() - t.length()) == 1) { |
38 | | - char[] longer = (s.length() > t.length()) ? schar : tchar; |
39 | | - char[] shorter = (longer == schar) ? tchar : schar; |
| 35 | + if (Math.abs(s.length() - t.length()) == 1) { |
| 36 | + char[] longer = (s.length() > t.length()) ? schar : tchar; |
| 37 | + char[] shorter = (longer == schar) ? tchar : schar; |
40 | 38 |
|
41 | | - int diffCnt = 0; |
42 | | - int i = 0; |
43 | | - int j = 0; |
44 | | - for (; i < shorter.length && j < longer.length; ) { |
45 | | - if (longer[j] != shorter[i]) { |
46 | | - diffCnt++; |
47 | | - j++; |
48 | | - } else { |
49 | | - i++; |
50 | | - j++; |
51 | | - } |
52 | | - } |
53 | | - return diffCnt == 1 |
54 | | - || diffCnt |
55 | | - == 0;//it could be the last char of the longer is the different one, in that case, diffCnt remains to be zero |
56 | | - } else if (s.length() == t.length()) { |
57 | | - int diffCnt = 0; |
58 | | - for (int i = 0; i < s.length(); i++) { |
59 | | - if (schar[i] != tchar[i]) { |
60 | | - diffCnt++; |
61 | | - } |
62 | | - if (diffCnt > 1) { |
| 39 | + int diffCnt = 0; |
| 40 | + int i = 0; |
| 41 | + int j = 0; |
| 42 | + for (; i < shorter.length && j < longer.length; ) { |
| 43 | + if (longer[j] != shorter[i]) { |
| 44 | + diffCnt++; |
| 45 | + j++; |
| 46 | + } else { |
| 47 | + i++; |
| 48 | + j++; |
| 49 | + } |
| 50 | + } |
| 51 | + return diffCnt == 1 |
| 52 | + || diffCnt |
| 53 | + == 0;//it could be the last char of the longer is the different one, in that case, diffCnt remains to be zero |
| 54 | + } else if (s.length() == t.length()) { |
| 55 | + int diffCnt = 0; |
| 56 | + for (int i = 0; i < s.length(); i++) { |
| 57 | + if (schar[i] != tchar[i]) { |
| 58 | + diffCnt++; |
| 59 | + } |
| 60 | + if (diffCnt > 1) { |
| 61 | + return false; |
| 62 | + } |
| 63 | + } |
| 64 | + return diffCnt == 1; |
| 65 | + } |
63 | 66 | return false; |
64 | | - } |
65 | 67 | } |
66 | | - return diffCnt == 1; |
67 | | - } |
68 | | - return false; |
69 | 68 | } |
70 | | - } |
71 | 69 | } |
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