Candy Crush

This question is about implementing a basic elimination algorithm for Candy Crush.

Given a 2D integer array board representing the grid of candy, different positive integers board[i][j] represent different types of candies. A value of board[i][j] = 0 represents that the cell at position (i, j) is empty. The given board represents the state of the game following the player's move. Now, you need to restore the board to a stable state by crushing candies according to the following rules:

1. If three or more candies of the same type are adjacent vertically or horizontally, "crush" them all at the same time - these positions become empty.

2. After crushing all candies simultaneously, if an empty space on the board has candies on top of itself, then these candies will drop until they hit a candy or bottom at the same time. (No new candies will drop outside the top boundary.)

3. After the above steps, there may exist more candies that can be crushed. If so, you need to repeat the above steps.

4. If there does not exist more candies that can be crushed (ie. the board is stable), then return the current board.

You need to perform the above rules until the board becomes stable, then return the current board.

My Solutions:

先处理每一行和每一列需要消掉的candy，再进行下落。只要有消除candy的动作，就要一直进行这个步骤。

public class Solution {
    /**
     * @param board: a 2D integer array
     * @return: the current board
     */
    public int[][] candyCrush(int[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) return board;
        boolean flag = true; // flag to indicate whether we need to proceed more
        while (flag) { 
            flag = processRowsAndCols(board);
            if (flag) dropCandies(board);
        }
        return board;
    }

    private boolean processRowsAndCols(int[][] board) {
        int rows = board.length, cols = board[0].length;
        boolean flag = false;

        // process rows
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols -2; j++) {
                int val = Math.abs(board[i][j]);
                // mark the cells to negative val if they can be crushed
                if (val != 0 && val == Math.abs(board[i][j + 1]) && val == Math.abs(board[i][j + 2])) {
                        board[i][j] = board[i][j + 1] = board[i][j + 2] = -val;
                        flag = true;
                }
            }
        }
        // process cols
        for (int j = 0; j < cols; j++) {
            for (int i = 0; i < rows - 2; i++) {
                int val = Math.abs(board[i][j]);
                if (val != 0 && val == Math.abs(board[i + 1][j]) && val == Math.abs(board[i + 2][j])) {
                    board[i][j] = board[i + 1][j] = board[i + 2][j] = -val;
                    flag = true;
                }
            }
        }
        return flag;
    }

    private void dropCandies(int[][] board) {
        int rows = board.length, cols = board[0].length;
        // 对每一列依次遍历
        for (int j = 0; j < cols; j++) {
            // 记录最下面一行，用来标记落下的正数需要更新到哪一行
            int bottomRow = rows - 1;
            // 对这一列的每一行，如果当前行的格子是正数，则需要下落（也可能就在同一行不变）
            // 如果是负数，说明当前格子会被覆盖掉，所以跳过
            for (int i = rows - 1; i >= 0; i--) {
                if (board[i][j] > 0) {
                    // 把正数的格子下落
                    board[bottomRow][j] = board[i][j];
                    // 更新标记
                    bottomRow--;
                }
            }
            // 在处理完这一列的每一行后，再把下落后上方的空位补成0
            while (bottomRow >= 0) {
                board[bottomRow][j] = 0;
                bottomRow--;
            }
        }
    }
}