Given two strings S and T, determine if they are both one edit distance apart.
class Solution {
public boolean isOneEditDistance(String s, String t) {
for (int i = 0; i < Math.min(s.length(), t.length()); i++) {
if (s.charAt(i) != t.charAt(i)) {
if (s.length() == t.length()) {
return s.substring(i + 1).equals(t.substring(i + 1));
} else if (s.length() < t.length()) {
return s.substring(i).equals(t.substring(i + 1));
} else {
return s.substring(i + 1).equals(t.substring(i));
}
}
}
return Math.abs(s.length() - t.length()) == 1;
}
}
时间复杂度O(N).