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`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*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
`board`

Will be in the range [3, 50].
- The length
`board[i]`

Will be in the range [3, 50].
- Each
`board[i][j]`

Will initially start as an integer in the range [1, 2000].

A two-dimensional array board is given to a question similar to the candy elimination game. The array represents candy and three or more consecutive identical numbers can be eliminated in the horizontal and vertical directions, after elimination, the number at the top fills in the space. Note that the candy that can be eliminated will be removed and then dropped for the second elimination. It remains stable until it is eliminated.

Solution: Mark all elements to be crush ed, remove these elements, and then drop the number above and change the vacancy to 0. The difficulty is how to determine which elements need to be 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