力扣 994. 腐烂的橘子

在给定的 m x n 网格 grid 中，每个单元格可以有以下三个值之一：

值 0 代表空单元格；
值 1 代表新鲜橘子；
值 2 代表腐烂的橘子。

每分钟，腐烂的橘子 周围 4 个方向上相邻 的新鲜橘子都会腐烂。

返回 直到单元格中没有新鲜橘子为止所必须经过的最小分钟数。如果不可能，返回 -1 。

解析：

广度优先遍历BFS

Java:

class Solution {
    public int orangesRotting(int[][] grid) {
        int M = grid.length, N = grid[0].length;
        Queue<int[]> queue = new LinkedList<>();
        int count = 0;
        for(int row = 0; row < M; row++) {
            for(int col = 0; col < N; col++) {
                if(grid[row][col] == 1) {
                    count++;
                } else if(grid[row][col] == 2) {
                    queue.add(new int[]{row, col});
                }
            }
        }
        int round = 0;
        while(count > 0 && !queue.isEmpty()) {
            round++;
            int n = queue.size();
            for(int i = 0; i < n; i++) {
                int[] orange = queue.poll();
                int row = orange[0];
                int col = orange[1];
                if(row - 1 >= 0 && grid[row - 1][col] == 1) {
                    grid[row - 1][col] = 2;
                    count--;
                    queue.add(new int[]{row - 1, col});
                }
                if(row + 1 < M && grid[row + 1][col] == 1) {
                    grid[row + 1][col] = 2;
                    count--;
                    queue.add(new int[]{row + 1, col});
                }
                if(col - 1 >= 0 && grid[row][col - 1] == 1) {
                    grid[row][col - 1] = 2;
                    count--;
                    queue.add(new int[]{row, col - 1});
                }
                if(col + 1 < N && grid[row][col + 1] == 1) {
                    grid[row][col + 1] = 2;
                    count--;
                    queue.add(new int[]{row, col + 1});
                }
            }
        }
        if(count > 0) {
            return -1;
        } else {
            return round;
        }
    }
}

JavaScript:

var orangesRotting = function(grid) {
    let count = 0, queue = [], total = 0, rot = 0;
    for (let i = 0; i < grid.length; i++) {
        for (let j = 0; j < grid[0].length; j++) {
            if (grid[i][j] == 2) {
                queue.push([i, j]);
                total++;
            }else if (grid[i][j] === 1) {
                total++;
            }
        }
    }
    if (total === 0) return 0;
    while (queue.length > 0 && rot < total) {
        let size = queue.length;
        rot += size;
        if (rot === total) return count;
        for (let i = 0; i < size; i++) {
            let [ti, tj] = queue.shift();
            if ((ti < grid.length - 1) && grid[ti + 1][tj] === 1) {
                grid[ti + 1][tj] = 2;
                queue.push([ti + 1, tj])
            }
            if (ti > 0 && grid[ti - 1][tj] === 1) {
                grid[ti - 1][tj] = 2;
                queue.push([ti - 1, tj]);
            }
            if ((tj < grid[0].length - 1) && grid[ti][tj + 1] === 1) {
                grid[ti][tj + 1] = 2;
                queue.push([ti, tj + 1]);
            }
            if (tj > 0 && grid[ti][tj - 1] === 1) {
                grid[ti][tj - 1] = 2;
                queue.push([ti, tj - 1]);
            }
        }
        count++;
    }
    return  -1;
};