723. Candy Crush

时间：2018-11-08 18:15:35 阅读：162 评论：0 收藏：0 [点我收藏+]

标签：about -- rom div xpl eset int class turn



 https://leetcode.com/problems/candy-crush/solution/


class Solution {
    public int[][] candyCrush(int[][] board) {
        int R = board.length, C = board[0].length;
        boolean todo = false; // reset 
        // horizontal scan
        for (int r = 0; r < R; ++r) {
            for (int c = 0; c + 2 < C; ++c) {
                int v = Math.abs(board[r][c]);
                if ( v != 0 && v == Math.abs(board[r][c+1]) && v == Math.abs(board[r][c+2])) {
                    board[r][c] = board[r][c+1] = board[r][c+2] = -v;
                    todo = true;
                }
            }
        }
        
        // vertical scan
        for (int r = 0; r + 2 < R; ++r) {
            for (int c = 0; c < C; ++c) {
                int v = Math.abs(board[r][c]);
                if ( v != 0 && v == Math.abs(board[r+1][c]) && v == Math.abs(board[r+2][c])) {
                    board[r][c] = board[r+1][c] = board[r+2][c] = -v;
                    todo = true;
                }
            }
        }
        
        // gravity steps 
        for (int c = 0; c < C; ++c) {
            int wr = R - 1; // write and read from the bottom, 
            for (int r = R-1; r >= 0; --r)
                // moves the wr to the first candy(negative number ) we see from the bottom, 
                // r moves all the time , until r is positive number , and we repalce wr with r value, and move wr up.  
                if (board[r][c] > 0)
                    board[wr--][c] = board[r][c];
            // after r reached 0, if wr is not reached 0 yet, we fill the wr with 0
            while (wr >= 0)
                board[wr--][c] = 0;
        }
        // check if we need to do this again 
        return todo ? candyCrush(board) : board;
    }
}

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:

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 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].

723. Candy Crush

标签：about -- rom div xpl eset int class turn

原文地址：https://www.cnblogs.com/tobeabetterpig/p/9929765.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行