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.
class Solution { public boolean isOneEditDistance(String s, String t) { if (Math.abs(s.length() - t.length()) > 1) { return false; } if (s.length() == t.length()) { return isOneModify(s, t); } else if (s.length() > t.length()) { return isOneDel(s, t); } else { return isOneDel(t, s); } } private boolean isOneModify(String s, String t) { int diff = 0, i = 0; while (i < s.length()) { if (s.charAt(i) != t.charAt(i)) { diff += 1; } i += 1; } return diff == 1; } private boolean isOneDel(String s, String t) { int i = 0; for (; i < s.length() && i < t.length(); i++) { if (s.charAt(i) != t.charAt(i)) { break; } } return s.substring(i + 1).equals(t.substring(i)); } }