【leetcode】994. Rotting Oranges

题目如下：

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 <= grid.length <= 10

1 <= grid[0].length <= 10

grid[i][j] is only 0, 1, or 2.

解题思路：采用BFS的思想，依次把坏的橘子附近的好的橘子变成坏的橘子，求出最大值。有一点要注意的是，有些好橘子四周都是空格的话，这个橘子不会变坏。因此最后要计算剩余好橘子的数量，以此确定是否返回-1。

代码如下：

class Solution(object):
    def orangesRotting(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        queue = []
        #visit = []
        fresh_count = 0
        for i in range(len(grid)):
            #visit.append([0]*len(grid[i]))
            for j in range(len(grid[i])):
                if grid[i][j] == 2:
                    queue.append((i,j,0))
                elif grid[i][j] == 1:
                    fresh_count += 1
        res = 0
        while len(queue) > 0:
            x,y,c = queue.pop(0)
            direction = [(0,1),(0,-1),(1,0),(-1,0)]
            for (i,j) in direction:
                if x + i >= 0 and x + i < len(grid) and y+j >=0 and y+j < len(grid[0]) and grid[x+i][y+j] == 1:
                    queue.append((x+i,y+j,c+1))
                    grid[x + i][y + j] = 2
                    res = max(res,c+1)
                    fresh_count -= 1
        if fresh_count > 0:
            return -1
        return res