原题链接在这里：https://leetcode.com/problems/candy-crush/
题目：
This question is about implementing a basic elimination algorithm for Candy Crush.
boardboard[i][j]board[i][j] = 0(i, j)stable state
- 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.
- stable), then return the current board.
You need to perform the above rules until the board becomes stable, then return the current board.
Example:
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:
boardboard[i]board[i][j]
题解：
Step 1: Mark 3 adjacent candies. Check if there are 3 adjacent candies. First check row by row, then column by column.
If there are, mark these values as negative.
Step 2: Crush them. Rewrite board with only positive numbers.
If there is crushing, that means there may be another round of crash, use recursion. Otherwise, there wouldn't be another round of crash, return the board.
Time Comlexity: O(M^2*n^2). m = board.length. n = board[0].length.
Each crash, there would be 3 crashed at minimum. Totally there are m*n candies. So recursion could run for m*n/3 times.
Space: O(1).
1 class Solution { 2 public int[][] candyCrush(int[][] board) { 3 if(board == null || board.length == 0 | board[0].length == 0){ 4 return board; 5 } 6 7 boolean todo = false; 8 int m = board.length; 9 int n = board[0].length; 10 for(int i = 0; i<m; i++){ 11 for(int j = 0; j<n-2; j++){ 12 int val = Math.abs(board[i][j]); 13 if(val!=0 && val==Math.abs(board[i][j+1]) && val==Math.abs(board[i][j+2])){ 14 todo = true; 15 board[i][j] = board[i][j+1] = board[i][j+2] = -val; 16 } 17 } 18 } 19 20 for(int j = 0; j<n; j++){ 21 for(int i = 0; i<m-2; i++){ 22 int val = Math.abs(board[i][j]); 23 if(val!=0 && val==Math.abs(board[i+1][j]) && val==Math.abs(board[i+2][j])){ 24 todo = true; 25 board[i][j] = board[i+1][j] = board[i+2][j] = -val; 26 } 27 } 28 } 29 30 for(int j = 0; j<n; j++){ 31 int br = m-1; 32 for(int i = m-1; i>=0; i--){ 33 if(board[i][j] > 0){ 34 board[br--][j] = board[i][j]; 35 } 36 } 37 38 while(br>=0){ 39 board[br--][j] = 0; 40 } 41 } 42 43 return todo ? candyCrush(board) : board; 44 } 45 } 来源：博客园
作者：Dylan_Java_NYC
链接：https://www.cnblogs.com/Dylan-Java-NYC/p/11421584.html