Monday, August 8, 2016

Google – Image And


Google – Image And
两个黑白image, 用2D matrix表示,matrix的边长为2^k。 要求设计一个数据结构来表示image,尽量做到space最少。然后对两个Image做And,黑黑得黑,白白得白,黑白得白。
[Solution]
看到2^k边长的条件,考点是QuadTree肯定错不了。
Quad tree既可以省空间,也可以省时间。因为如果某一大块下面所有的cell都是黑(或者都是白),那么它就不需要任何sub tree了。做And的时候也可以直接返回,要比O(n^2) Brute force一个cell一个cell的做And快很多。
class TNode {
  public char val;
 
  public TNode leftTop;
  public TNode rightTop;
  public TNode leftBot;
  public TNode rightBot;
 
  public TNode(char val) {
    this.val = val;
  }
}
 
class Solution {
 
  public TNode buildTree(char[][] image) {
    int n = image.length;
    return buildTree(image, 0, 0, n - 1, n - 1);
  }
 
  public TNode buildTree(char[][] image, int row1, int col1, int row2, int col2) {
    if (row1 < 0 || row1 >= image.length || row2 < 0 || row2 >= image.length || col1 < 0 || col1 >= image.length || col2 < 0 || col2 >= image.length || row1 > row2 || col1 > col2) {
      return null;
    }
 
    if (row1 == row2 && col1 == col2) {
      TNode result = new TNode(image[row1][col1]);
      if (image[row1][col1] == '1') {
        result.val = '1';
      }
      else {
        result.val = '0';
      }
      return result;
    }
 
    int midX = row1 + (row2 - row1) / 2;
    int midY = col1 + (col2 - col1) / 2;
 
    TNode root = new TNode('#');
 
    TNode leftTop = buildTree(image, row1, col1, midX, midY);
    TNode rightTop = buildTree(image, row1, midY + 1, midX, col2);
    TNode leftBot = buildTree(image, midX + 1, col1, row2, midY);
    TNode rightBot = buildTree(image, midX + 1, midY + 1, row2, col2);
 
    if (leftTop.val == '1'
        && rightTop.val == '1'
        && leftBot.val == '1'
        && rightBot.val == '1') {
 
      root.val = '1';
    }
    else if (leftTop.val == '0'
        && rightTop.val == '0'
        && leftBot.val == '0'
        && rightBot.val == '0') {
 
      root.val = '0';
    }
    else {
      root.leftTop = leftTop;
      root.rightTop = rightTop;
      root.leftBot = leftBot;
      root.rightBot = rightBot;
    }
 
    return root;
  }
 
  public TNode imageAnd(TNode root1, TNode root2) {
    TNode root = new TNode('#');
    if (root1 == null && root2 == null) {
      return null;
    }
    if (root1.val == '1' && root2.val == '1') {
      return new TNode('1');
    }
    else if (root1.val == '0' || root2.val == '0') {
      return new TNode('0');
    }
    else if (root1.val == '#' && root2.val == '1') {
      root.leftTop = imageAnd(root1.leftTop, root2);
      root.rightTop = imageAnd(root1.rightTop, root2);
      root.leftBot = imageAnd(root1.leftBot, root2);
      root.rightBot = imageAnd(root1.rightBot, root2);
    }
    else if (root1.val == '1' && root2.val == '#') {
      root.leftTop = imageAnd(root1, root2.leftTop);
      root.rightTop = imageAnd(root1, root2.rightTop);
      root.leftBot = imageAnd(root1, root2.leftBot);
      root.rightBot = imageAnd(root1, root2.rightBot);
    }
    else {
      root.leftTop = imageAnd(root1.leftTop, root2.leftTop);
      root.rightTop = imageAnd(root1.rightTop, root2.rightTop);
      root.leftBot = imageAnd(root1.leftBot, root2.leftBot);
      root.rightBot = imageAnd(root1.rightBot, root2.rightBot);
    }
    return root;
  }
 
  public void toMatrix(TNode root, char[][] matrix, int row1, int col1, int row2, int col2) {
    if (root == null) {
      return;
    }
 
    if (root.val == '1' || root.val == '0') {
      for (int i = row1; i <= row2; i++) {
        for (int j = col1; j <= col2; j++) {
          matrix[i][j] = root.val;
        }
      }
    }
    else {
      int midX = row1 + (row2 - row1) / 2;
      int midY = col1 + (col2 - col1) / 2;
      toMatrix(root.leftTop, matrix, row1, col1, midX, midY);
      toMatrix(root.rightTop, matrix, row1, midY + 1, midX, col2);
      toMatrix(root.leftBot, matrix, midX + 1, col1, row2, midY);
      toMatrix(root.rightBot, matrix, midX + 1, midY + 1, row2, col2);
    }
  }
}
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