题目：
Given two strings S and T, determine if they are both one edit distance apart.
链接： http://leetcode.com/problems/one-edit-distance/
6/14/2017
2ms, 58%
提交好几次才通过，所以代码像打补丁一样。注意的test case：
2个空string -- false
相同的string -- false
当检查出来不一致时，根据长度来比较之后的substring
1 public class Solution { 2 public boolean isOneEditDistance(String s, String t) { 3 if (s == null || t == null) { 4 return false; 5 } 6 int sLen = s.length(); 7 int tLen = t.length(); 8 if (sLen == 0 && tLen == 0) { 9 return false; 10 } 11 if (sLen == 0 && tLen == 1 || sLen == 1 && tLen == 0) { 12 return true; 13 } 14 if (Math.abs(sLen - tLen) > 1) { 15 return false; 16 } 17 int i = 0, j = 0; 18 while (i < sLen && j < tLen) { 19 char c = s.charAt(i); 20 char d = t.charAt(j); 21 if (c != d) { 22 break; 23 } 24 i++; 25 j++; 26 } 27 if (i == sLen && j == tLen) { 28 return false; 29 } 30 if (sLen == tLen) { 31 return s.substring(i + 1, sLen).equals(t.substring(j + 1, tLen)); 32 } else if (sLen < tLen) { 33 return s.substring(i, sLen).equals(t.substring(j + 1, tLen)); 34 } else { 35 return s.substring(i + 1, sLen).equals(t.substring(j, tLen)); 36 } 37 } 38 }
别人的做法，
跟我类似，也是利用了后面的substring，不过前期省很多
https://discuss.leetcode.com/topic/30308/my-clear-java-solution-with-explanation
1 public boolean isOneEditDistance(String s, String t) { 2 for (int i = 0; i < Math.min(s.length(), t.length()); i++) { 3 if (s.charAt(i) != t.charAt(i)) { 4 if (s.length() == t.length()) // s has the same length as t, so the only possibility is replacing one char in s and t 5 return s.substring(i + 1).equals(t.substring(i + 1)); 6 else if (s.length() < t.length()) // t is longer than s, so the only possibility is deleting one char from t 7 return s.substring(i).equals(t.substring(i + 1)); 8 else // s is longer than t, so the only possibility is deleting one char from s 9 return t.substring(i).equals(s.substring(i + 1)); 10 } 11 } 12 //All previous chars are the same, the only possibility is deleting the end char in the longer one of s and t 13 return Math.abs(s.length() - t.length()) == 1; 14 }
另外一种
1 public class Solution { 2 public boolean isOneEditDistance(String s, String t) { 3 if (s == null || t == null || s.equals(t) || Math.abs(s.length() - t.length()) > 1) return false; 4 if (s.length() > t.length()) return isOneEditDistance(t, s); 5 boolean hasDiff = false; 6 for (int i = 0, j = 0; i < s.length(); i++, j++) { 7 if (s.charAt(i) != t.charAt(j)) { 8 if (hasDiff) return false; 9 hasDiff = true; 10 if (s.length() < t.length()) i--; 11 } 12 } 13 return true; 14 } 15 }
2个指针
https://discuss.leetcode.com/topic/27379/java-python-two-pointer-solution
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