- Difficulty: Medium
Given two strings S and T, determine if they are both one edit distance apart.
Company:
思路：一共有3种情况满足1edit distance
case1：len1 = len2, "abc" vs "adc"//replace one char from str1
1: check if diff > 1
case2: len1 - len2 = 1,"abc" vs "ac" //delete one char from str1
case3: len2 - len1 = 1, "ab" vs "abc";//delete one char from str2
2.3 find the different char, compare the rest of substring
关键字： substring, str.equals(str2);
public class Solution { public boolean isOneEditDistance(String s, String t) { int count =0; int len1 = s.length(), len2 = t.length(); if(len1 == len2){ for(int i = 0; i < s.length(); i++){ if(s.charAt(i) != t.charAt(i)) count++; if(count > 1) return false; } return !s.equals(t); }else if(len1 - len2 == 1||len1 - len2 == -1){ int i = 0, j = 0; while(i < len1 && j < len2){ if(s.charAt(i) != t.charAt(j)){ String tmp1, tmp2; if(len1 < len2){ tmp1 = s.substring(i, len1); tmp2 = t.substring(j + 1, len2); }else{ tmp1 = s.substring(i + 1, len1); tmp2 = t.substring(j, len2); } return tmp1.equals(tmp2); } i++; j++; } return true;//"a", "" } return false; } }
