723. 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:
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.
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.)
After the above steps, there may exist more candies that can be crushed. If so, you need to repeat the above steps.
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.
Example 1:
Input:
board = 
[[110,5,112,113,114],[210,211,5,213,214],[310,311,3,313,314],[410,411,412,5,414],[5,1,512,3,3],[610,4,1,613,614],[710,1,2,713,714],[810,1,2,1,1],[1,1,2,2,2],[4,1,4,4,1014]]
Output:
[[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[110,0,0,0,114],[210,0,0,0,214],[310,0,0,113,314],[410,0,0,213,414],[610,211,112,313,614],[710,311,412,613,714],[810,411,512,713,1014]]
Explanation: 
Note:
The length of board will be in the range [3, 50].
The length of board[i] will be in the range [3, 50].
Each board[i][j] will initially start as an integer in the range [1, 2000].
题意解释：
一次性消除board中所有可以消除的糖果，然后才下落，形成新的糖果。

思路:
标记出所有需要被crash 掉的元素
用一个data structure去记录可以被消除的糖果的位置坐标 （检查横向、纵向相同糖果的个数，只要有一个方向有三个以上相同糖果，当前这个糖果就能被删除）
将这些元素设置为0，并且crash
crash 处理：
    实际上相当于two pointers把0给移到board的顶部，设置两个pointers从board的尾部开始往上走。假如值是0， 快指针往前走；假如值非0，快指针和慢指针交换value, 慢指针和快指针都向前走一位

模拟crash 的过程：
6          6           6        6         0
4          4           4        0         0
3          3           0        0         0
0   =>   0    =>  0   => 0  =>   6
1          0           0        4         4
0          0           3        3         3
0          1           1        1         1
Time: O((row * col)^2)
Space: O(1)

Solution (Recursion):
Space: call stack
class CandyCrushGame {
    class Point{
        int x;
        int y;
        public Point(int x, int y ) {
            this.x = x;
            this.y = y;
        }
    }
    public int[][] candyCrush(int[][] board) {
        Set<Point> markDeleted = new HashSet<>();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == 0) continue;
                if ((i - 2 >= 0 && board[i][j] == board[i- 1][j] && board[i][j] == board[i - 2][j])||
                   (j - 2 >= 0 && board[i][j] == board[i][j - 1] && board[i][j] == board[i][j - 2])||
                   (i + 2 < board.length && board[i][j] == board[i + 1][j] && board[i][j] == board[i + 2][j])||
                   (j + 2 < board[0].length && board[i][j] == board[i][j + 1] && board[i][j] == board[i][j + 2])||
                   (i - 1 >= 0 && i + 1 < board.length && board[i - 1][j] == board[i][j] && board[i][j] == board[i + 1][j])||
                   (j - 1 >= 0 && j + 1 < board[0].length && board[i][j - 1] == board[i][j] && board[i][j] == board[i][j + 1])) {
                    markDeleted.add(new Point(i, j));
                }
            }
        }
        if (markDeleted.isEmpty()) return board;
        for(Point p: markDeleted) {
            board[p.x][p.y] = 0;
        }
        
        dropBoard(board);
        return candyCrush(board);
    }
    
    private void dropBoard(int[][] board) {
        for (int j = 0; j < board[0].length; j++) {
            int bot = board.length - 1;
            int top = board.length - 1;
            while (top >= 0) {
                if (board[top][j] == 0) {
                    top--;
                }
                else {
                    board[bot--][j] = board[top--][j];
                }
            }
            while (bot >= 0) {
                board[bot--][j] = 0;
            }
        }
    }
}
Solution (Iterative):
class Point {
        int x;
        int y;
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
    public int[][] candyCrush(int[][] board) {
        int m = board.length;
        int n = board[0].length;
        
        while (true) {
            List<Point> deletion = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (board[i][j] == 0) continue;
                    int x0 = i;
                    int x1 = i;
                    int y0 = j;
                    int y1 = j;
                    while (x0 >= 0 && x0 > i - 3 && board[x0][j] == board[i][j]) --x0;
                    while (x1 < m && x1 < i + 3 && board[x1][j] == board[i][j]) ++x1;
                    while (y0 >= 0 && y0 > j - 3 && board[i][y0] == board[i][j]) --y0;
                    while (y1 < n && y1 < j + 3 && board[i][y1] == board[i][j]) ++y1;
                    if (x1 - x0 > 3 || y1 - y0 > 3) deletion.add(new Point(i, j));
                }
            }
            if (deletion.size() == 0) break;
            for (Point p: deletion) {
                board[p.x][p.y] = 0;
            }
            for (int j = 0; j < n; j++) {
                int t = m - 1;
                for (int i = m - 1; i >= 0; i--) {
                    if (board[i][j] != 0) {
                        int tmp = board[t][j];
                        board[t][j] = board[i][j];
                        board[i][j] = tmp;
                        t--;
                    }
                }
            }
        }
        return board;
    }