994. Rotting Oranges - Medium

In a given grid, each cell can have one of three values:

• the value 0 representing an empty cell;
• the value 1 representing a fresh orange;
• the value 2 representing a rotten orange.

Every minute, any fresh orange that is adjacent (4-directionally) to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange.  If this is impossible, return -1 instead.

Example 1:

Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4

Example 2:

Input: [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation:  The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.

Example 3:

Input: [[0,2]]
Output: 0
Explanation:  Since there are already no fresh oranges at minute 0, the answer is just 0.

Note:

1. 1 <= grid.length <= 10
2. 1 <= grid[0].length <= 10
3. grid[i][j] is only 0, 1, or 2.

BFS

time = O(m * n), space = O(m * n)

class Solution {
    int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    
    public int orangesRotting(int[][] grid) {
        if(grid == null || grid.length == 0) {
            return 0;
        }
        int freshCount = 0;
        Queue<int[]> q = new LinkedList<>();
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(grid[i][j] == 2) {
                    q.offer(new int[] {i, j});
                } else if(grid[i][j] == 1) {
                    freshCount++;
                }
            }
        }
        
        if(freshCount == 0) {
            return 0;
        }
        
        int time = 0;
        while(!q.isEmpty()) {
            int size = q.size();
            for(int i = 0; i < size; i++) {
                int[] cur = q.poll();
                for(int[] dir : dirs) {
                    int x = cur[0] + dir[0];
                    int y = cur[1] + dir[1];
                    if(x >= 0 && x < grid.length && y >= 0 && y < grid[0].length && grid[x][y] == 1) {
                        grid[x][y] = 2;
                        q.offer(new int[] {x, y});
                        freshCount--;
                    }
                }
            }
            time++;
        }
        
        return freshCount == 0 ? time - 1 : -1;
    }
}

标签：994,Medium,Rotting,++,int,length,freshCount,grid,orange
来源： https://www.cnblogs.com/fatttcat/p/13473355.html