all possible patterns for N russian squares

标签：interview

Question:

A traditional russion square is made by 4 blocks, and has 7 different patterns.

Given N blocks, how many different patterns would be.

N = 1, P# = 1

N = 2, P# = 1

N = 3, P# = 3

N = 4, P# = 7.

Assume N = k, P# = Pk

then when N = k+1, P#  = for each P in Pk, append one more block around P, remove dups.

class Pattern
{
  List<Block> blocks;
  
  // Return true if this pattern can be translated to another patter p by rotate or moving.
  boolean isSame(Pattern p)
  {
    // TODO
  }  
}

List<Pattern> allPatthens(n)
{
  if (n == 1)
  {
    return Pattern.builder().build(); 
  }
  
  List<Pattern> patternsForNMinusOne = numOfPattern(n - 1);
  List<Pattern> toReturn = new ArrayList<>();
  for (Pattern p : patternsForNMinus1)
  {
    List<Pattern> patternsWithOneMoreBlock = patternsWithOneMoreBlock(p);
    for (Pattern newP : patternsWithOneMoreBlock)
    {
      if (!contains(toReturn, newP))
      {
        toReturn.add(newP);
      }
    }
  }
  
  return toReturn;
}


boolean contains(List<Pattern> list, Pattern p)
{
  for (Pattern pInList : list)
  {
    if (pInList.isSame(p))
      return true;
  }
  return false;
}


// Given pattern p.
// Return all unique valid patterns with one more block.
List<Pattern> patternsWithOneMoreBlock(Pattern p)
{
  List<Pattern> toReturn = new ArrayList<>();
  for (Block b : p.blocks())
  {
    List<Block> newBlocks = Arrays.as(b.left(), b.right(), b.up(), b.down());
    for (Block newBlock : newBlocks)
    {
      if (!p.contains(newBlock))
      {
        // A new pattern found
        Pattern newP = p.appendBlock(newBlock);
        if (!contains(toReturn, newP))
        {
          toReturn.add(newP);
        }
      }
    }
  }
  return toReturn;
}


// Too complicated.

all possible patterns for N russian squares

标签：interview

原文地址：http://7371901.blog.51cto.com/7361901/1589958