One Edit Distance

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

 

 

注意這道題：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面經有一道比較狠的這個題的變形：

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

這題就是one edit distance的變形題，難點在於給的Iterator，事先不知道兩個file
的長度，也不容許用extra space（因此不能轉成兩個string再比），那麼一個個往前
跑的時候就得三種狀況都考慮。。。。

個人作法

 

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;
          }
      }
}