leetcode1162 As Far from Land as Possible

 """
 Given an N x N grid containing only values  and , where  represents water and  represents land, find a water cell such that its distance to the nearest land cell is maximized and return the distance.
 The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|.
 If no land or water exists in the grid, return -.
 Example :
 Input: [[,,],[,,],[,,]]
 Output:
 Explanation:
 The cell (, ) is as far as possible from all the land with distance .
 Example :
 Input: [[,,],[,,],[,,]]
 Output:
 Explanation:
 The cell (, ) is as far as possible from all the land with distance .
 """
 """
 BFS的方法
 先找到grid中所有的1的位置，保存在queue中，若全是1或者没有1，返回-
 从这些1所在的位置开始进行广度优先搜索，即搜索四个方向，
 如果搜索的位置上的值为0，保存这些坐标，
 作为下一次搜索的起点，并将这些位置的值设置为1，以免重复搜索。
 这样每扩散一层，计数器加1，直到没有可搜索的起点，返回计数器的值就为最远的距离
 """
 class Solution:
     def maxDistance(self, grid):
         row, col = len(grid), len(grid[]) #求出地图的行列
         queue = [] #保存地图中出现''的位置
         for i in range(row):    #遍历地图。将为''的所有位置放入队列
             for j in range(col):
                 if grid[i][j] == :
                     queue.append((i, j))
         if len(queue) == row*col or not queue: #如果地图全为''或者全为''，返回-
             return -
         level =    #记录每次扩张的距离
         while queue:
             newqueue = []  #记录每层扩张
             for x, y in queue:
                 for i, j in [[x-, y], [x+, y], [x, y-], [x, y+]]: #每层扩张的四个方向
                     if  <= i <row and <= j < col and grid[i][j] == :
                         newqueue.append((i, j))  #如果为0，搜索到了，保存位置放入队列，下一轮搜索
                         grid[i][j] =     #并将搜索过的位置 变为1
             queue = newqueue
             level +=
         return (level-)