723. Candy Crush

1723. Candy Crush
2This question is about implementing a basic elimination algorithm for Candy Crush.
3Given 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:
4If three or more candies of the same type are adjacent vertically or horizontally, "crush" them all at the same time - these positions become empty.
5After 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.)
6After the above steps, there may exist more candies that can be crushed. If so, you need to repeat the above steps.
7If there does not exist more candies that can be crushed (ie. the board is stable), then return the current board.
8You need to perform the above rules until the board becomes stable, then return the current board.
9Example 1:
10Input:
11board = 
12[[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]]
13Output:
14[[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]]
15Explanation: 
16Note:
17The length of board will be in the range [3, 50].
18The length of board[i] will be in the range [3, 50].
19Each board[i][j] will initially start as an integer in the range [1, 2000].
20题意解释：
21一次性消除board中所有可以消除的糖果，然后才下落，形成新的糖果。
22
23思路:
24标记出所有需要被crash 掉的元素
25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）
26将这些元素设置为0，并且crash
27crash 处理：
28    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
29
30模拟crash 的过程：
316          6           6        6         0
324          4           4        0         0
333          3           0        0         0
340   =>   0    =>  0   => 0  =>   6
351          0           0        4         4
360          0           3        3         3
370          1           1        1         1
38Time: O((row * col)^2)
39Space: O(1)
40
41Solution (Recursion):
42Space: call stack
43class CandyCrushGame {
44    class Point{
45        int x;
46        int y;
47        public Point(int x, int y ) {
48            this.x = x;
49            this.y = y;
50        }
51    }
52    public int[][] candyCrush(int[][] board) {
53        Set<Point> markDeleted = new HashSet<>();
54        for (int i = 0; i < board.length; i++) {
55            for (int j = 0; j < board[0].length; j++) {
56                if (board[i][j] == 0) continue;
57                if ((i - 2 >= 0 && board[i][j] == board[i- 1][j] && board[i][j] == board[i - 2][j])||
58                   (j - 2 >= 0 && board[i][j] == board[i][j - 1] && board[i][j] == board[i][j - 2])||
59                   (i + 2 < board.length && board[i][j] == board[i + 1][j] && board[i][j] == board[i + 2][j])||
60                   (j + 2 < board[0].length && board[i][j] == board[i][j + 1] && board[i][j] == board[i][j + 2])||
61                   (i - 1 >= 0 && i + 1 < board.length && board[i - 1][j] == board[i][j] && board[i][j] == board[i + 1][j])||
62                   (j - 1 >= 0 && j + 1 < board[0].length && board[i][j - 1] == board[i][j] && board[i][j] == board[i][j + 1])) {
63                    markDeleted.add(new Point(i, j));
64                }
65            }
66        }
67        if (markDeleted.isEmpty()) return board;
68        for(Point p: markDeleted) {
69            board[p.x][p.y] = 0;
70        }
71        
72        dropBoard(board);
73        return candyCrush(board);
74    }
75    
76    private void dropBoard(int[][] board) {
77        for (int j = 0; j < board[0].length; j++) {
78            int bot = board.length - 1;
79            int top = board.length - 1;
80            while (top >= 0) {
81                if (board[top][j] == 0) {
82                    top--;
83                }
84                else {
85                    board[bot--][j] = board[top--][j];
86                }
87            }
88            while (bot >= 0) {
89                board[bot--][j] = 0;
90            }
91        }
92    }
93}
94Solution (Iterative):
95class Point {
96        int x;
97        int y;
98        public Point(int x, int y) {
99            this.x = x;
100            this.y = y;
101        }
102    }
103    public int[][] candyCrush(int[][] board) {
104        int m = board.length;
105        int n = board[0].length;
106        
107        while (true) {
108            List<Point> deletion = new ArrayList<>();
109            for (int i = 0; i < m; i++) {
110                for (int j = 0; j < n; j++) {
111                    if (board[i][j] == 0) continue;
112                    int x0 = i;
113                    int x1 = i;
114                    int y0 = j;
115                    int y1 = j;
116                    while (x0 >= 0 && x0 > i - 3 && board[x0][j] == board[i][j]) --x0;
117                    while (x1 < m && x1 < i + 3 && board[x1][j] == board[i][j]) ++x1;
118                    while (y0 >= 0 && y0 > j - 3 && board[i][y0] == board[i][j]) --y0;
119                    while (y1 < n && y1 < j + 3 && board[i][y1] == board[i][j]) ++y1;
120                    if (x1 - x0 > 3 || y1 - y0 > 3) deletion.add(new Point(i, j));
121                }
122            }
123            if (deletion.size() == 0) break;
124            for (Point p: deletion) {
125                board[p.x][p.y] = 0;
126            }
127            for (int j = 0; j < n; j++) {
128                int t = m - 1;
129                for (int i = m - 1; i >= 0; i--) {
130                    if (board[i][j] != 0) {
131                        int tmp = board[t][j];
132                        board[t][j] = board[i][j];
133                        board[i][j] = tmp;
134                        t--;
135                    }
136                }
137            }
138        }
139        return board;
140    }