# LWC 57：723. Candy Crush

## LWC 57：723. Candy Crush

Problem:

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: [[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].

```    public int[][] candyCrush(int[][] board) {
int n = board.length;
int m = board[0].length;
boolean[][] next = new boolean[n][m];
while (getNext(next, board)){
List<Integer>[] ans = new ArrayList[m];
for (int i = 0; i < m; ++i){
ans[i] = new ArrayList<>();
for (int j = n - 1; j >= 0; --j){
}
}
board = new int[n][m];
// 重写
for (int j = 0; j < m; ++j){
for (int i = 0; i < ans[j].size(); ++i){
board[n - 1 - i][j] = ans[j].get(i);
}
}
next = new boolean[n][m];
}
return board;
}

public boolean getNext(boolean[][] next, int[][] board){
boolean exist = false;
// 遍历行
int n = board.length;
int m = board[0].length;

for (int i = 0; i < n; ++i){
int prev = board[i][0];
int count = 1;
for (int j = 1; j < m; ++j){
if (board[i][j] == prev && board[i][j] != 0){
count ++;
}
else{
if (count >= 3){
for (int l = 0; l < count; ++l){
next[i][j - 1 - l] = true;
}
exist = true;
}
count = 1;
}
prev = board[i][j];
}

if (count >= 3){
for (int l = 0; l < count; ++l){
next[i][m - 1 - l] = true;
}
exist = true;
}
}

// 遍历列
for (int i = 0; i < m; ++i){
int prev = board[0][i];
int count = 1;
for (int j = 1; j < n; ++j){
if (board[j][i] == prev && board[j][i] != 0){
count ++;
}
else{
if (count >= 3){
for (int l = 0; l < count; ++l){
next[j - 1 - l][i] = true;
}
exist = true;
}
count = 1;
}
prev = board[j][i];
}

if (count >= 3){
for (int l = 0; l < count; ++l){
next[n - 1 - l][i] = true;
}
exist = true;
}
}

return exist;
}    ```

0 条评论

## 相关文章

### Mix 10 上的asp.net mvc 2的相关Session

Beyond File | New Company: From Cheesy Sample to Social Platform Scott Hansel...

2787

5184

### js登录滑动验证，不滑动无法登陆

js的判断这里是根据滑块的位置进行判断，应该是用一个flag判断 <%@ page language="java" contentType="text/html...

8868

### Kit 3D 更新

Kit3D is a 3D graphics engine written for Microsoft Silverlight. Kit3D was inita...

2946

### Miguel de Icaza 细说 Mix 07大会上的Silverlight和DLR

Mono之父Miguel de Icaza 详细报道微软Mix 07大会上的Silverlight和DLR ,上面还谈到了Mono and Silverligh...

3017

4224

4425

5248

### Luminous版本PG 分布调优

Luminous版本开始新增的balancer模块在PG分布优化方面效果非常明显，操作也非常简便，强烈推荐各位在集群上线之前进行这一操作，能够极大的提升整个集群...

3685

### cocos2dx 打灰机

#include "GamePlane.h" #include "PlaneSprite.h" #include "BulletNode.h" #include...

7336