One Edit Distance

Given two strings S and T, determine if they are both one edit distance apart.

注意这道题：Must be exactly one distance apart. Not the same.

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 t.substring(i).equals(s.substring(i+1));
            }
        }
        
        //Corner case, last char
        return Math.abs(s.length() - t.length()) == 1;
    }
}

　　另外FB面经有一道比较狠的这个题的变形：

class IntFileIterator {
  boolean hasNext();
  int next();
}

class{
  public boolean isDistanceZeroOrOne(IntFileIterator a, IntFileIterator b)；

}
// return if the distance between a and b is at most 1.. 
// Distance: minimum number of modifications to make a=b
// Modification:
//   1. change an int in a
//   2. insert an int to a
//   3. remove an int from a

这题就是one edit distance的变形题，难点在于给的Iterator，事先不知道两个file
的长度，也不允许用extra space（所以不能转成两个string再比），那么一个个往前
跑的时候就得三种情况都考虑。。。。

我的做法

public class Solution {
    class IntFileIterator {
        boolean hasNext();
        int next();
    }
    
    public boolean isDistanceZeroOrOne(IntFileIterator a, IntFileIterator b) {
        return check(a, b, 0);
    }
  
      public boolean check(IntFileIterator a, IntFileIterator b, int distance){
          IntFileIterator aa = new InFileIterator(a); // copy of iterator a before next() function
          IntFileIterator bb = new InFileIterator(b);
          while (a.hasNext() && b.hasNext()) {
              int s = a.next();
              int t = b.next();
              if(s != t){
                  IntFileIterator aaa = new InFileIterator(a); //copy of iterator a after next() function
                  IntFileIterator bbb = new InFileIterator(b);
                  distance++;
                  if(distance>1) return false;
                  return check(aa, b, distance) || check(a, bb, distance) || check(aaa, bbb, distance);
              }
              else{
                  return check(a, b, distance);
              }
          }
          
          if(distance == 1){
              return !a.hasNext() && !b.hasNext();
          }else { //(distance ==0)
              IntFileIterator k = a.hasNext()? a : b;
              int count = 0;
              while (k.hasNext()) {
                  k.next();
                  count++;
              }
              return count<=1;
          }
      }
}