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

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.

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:

1. The length of board will be in the range [3, 50].
2. The length of board[i] will be in the range [3, 50].
3. Each board[i][j] will initially start as an integer in the range [1, 2000].

一個類似糖果消消樂遊戲的題，給一個二維數組board，數組代表糖果，在水平和豎直方向上3個或以上連續相同的數字可以被消除，消除後位於上方的數字會填充空位。註意是所以能消除的糖果消除後然後再落下，進行第二次消除。直到最後沒有能消除的了，達到穩定狀態。

解法：標記出所有需要被crush的元素，然後去掉這些元素，將上面的數下降，空位變為0。難的地方是如何確定哪些元素需要被crush。

Python:

class Solution(object):
    def candyCrush(self, board):
        """
        :type board: List[List[int]]
        :rtype: List[List[int]]
        """
        R, C = len(board), len(board[0])
        changed = True

        while changed:
            changed = False

            for r in xrange(R):
                for c in xrange(C-2):
                    if abs(board[r][c]) == abs(board[r][c+1]) == abs(board[r][c+2]) != 0:
                        board[r][c] = board[r][c+1] = board[r][c+2] = -abs(board[r][c])
                        changed = True

            for r in xrange(R-2):
                for c in xrange(C):
                    if abs(board[r][c]) == abs(board[r+1][c]) == abs(board[r+2][c]) != 0:
                        board[r][c] = board[r+1][c] = board[r+2][c] = -abs(board[r][c])
                        changed = True

            for c in xrange(C):
                i = R-1
                for r in reversed(xrange(R)):
                    if board[r][c] > 0:
                        board[i][c] = board[r][c]
                        i -= 1
                for r in reversed(xrange(i+1)):
                    board[r][c] = 0

        return board　

C++:

// Time:  O((R * C)^2)
// Space: O(1)

class Solution {
public:
    vector<vector<int>> candyCrush(vector<vector<int>>& board) {
        const auto R = board.size(), C = board[0].size();
        bool changed = true;
        
        while (changed) {
            changed = false;
            
            for (int r = 0; r < R; ++r) {
                for (int c = 0; c + 2 < C; ++c) {
                    auto v = abs(board[r][c]);
                    if (v != 0 && v == abs(board[r][c + 1]) && v == abs(board[r][c + 2])) {
                        board[r][c] = board[r][c + 1] = board[r][c + 2] = -v;
                        changed = true;
                    }
                }
            }
            
            for (int r = 0; r + 2 < R; ++r) {
                for (int c = 0; c < C; ++c) {
                    auto v = abs(board[r][c]);
                    if (v != 0 && v == abs(board[r + 1][c]) && v == abs(board[r + 2][c])) {
                        board[r][c] = board[r + 1][c] = board[r + 2][c] = -v;
                        changed = true;
                    }
                }
            }

            for (int c = 0; c < C; ++c) {
                int empty_r = R - 1;
                for (int r = R - 1; r >= 0; --r) {
                    if (board[r][c] > 0) {
                        board[empty_r--][c] = board[r][c];
                    }
                }
                for (int r = empty_r; r >= 0; --r) {
                    board[r][c] = 0;
                }
            }
        }
        
        return board;
    }
};　　

C++:

class Solution {
public:
    vector<vector<int>> candyCrush(vector<vector<int>>& board) {
        int m = board.size(), n = board[0].size();
        while (true) {
            vector<pair<int, int>> del;
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (board[i][j] == 0) continue;
                    int x0 = i, x1 = i, y0 = j, 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) del.push_back({i, j});
                }
            }
            if (del.empty()) break;
            for (auto a : del) board[a.first][a.second] = 0;
            for (int j = 0; j < n; ++j) {
                int t = m - 1;
                for (int i = m - 1; i >= 0; --i) {
                    if (board[i][j]) swap(board[t--][j], board[i][j]);   
                }
            }
        }
        return board;
    }
};

C++:

class Solution {
public:
    vector<vector<int>> candyCrush(vector<vector<int>>& board) {
        int m = board.size(), n = board[0].size();
        bool toBeContinued = false;
        
        for (int i = 0; i < m; ++i) { // horizontal crushing 
            for (int j = 0; j + 2 < n; ++j) {
                int& v1 = board[i][j];
                int& v2 = board[i][j+1];
                int& v3 = board[i][j+2];
                
                int v0 = std::abs(v1);
                
                if (v0 && v0 == std::abs(v2) && v0 == std::abs(v3)) {
                    v1 = v2 = v3 = - v0;
                    toBeContinued =  true;
                }
            }
        }
        
        for (int i = 0; i + 2 < m; ++i) {  // vertical crushing
            for (int j = 0; j < n; ++j) {
                int& v1 = board[i][j];
                int& v2 = board[i+1][j];
                int& v3 = board[i+2][j];
                int v0 = std::abs(v1);
                if (v0 && v0 == std::abs(v2) && v0 == std::abs(v3)) {
                    v1 = v2 = v3 = -v0;
                    toBeContinued = true;
                }
            }
        }
        
        for (int j = 0; j < n; ++j) { // gravity step
            int dropTo = m - 1;
            for (int i = m - 1; i >= 0; --i) {
                if (board[i][j] >= 0) {
                    board[dropTo--][j] = board[i][j];
                }
            }
            
            for (int i = dropTo; i >= 0; i--) {
                board[i][j] = 0;
            }
        }
        
        return toBeContinued ? candyCrush(board) : board;
        
    }
};

[LeetCode] 723. Candy Crush 糖果粉碎