[LeetCode] 723. Candy Crush 糖果粉碎

2021年09月15日 阅读数：1

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

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:python

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.数组

Example 1:ide

```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:post

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].

Python:rest

```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++:code

```// 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++:orm

```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;
}
}

}
};
```