leetcode1162 As Far from Land as Possible

"""
Given an N x N grid containing only values 0 and 1, where 0 represents water and 1 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 -1.
Example 1:
Input: [[1,0,1],[0,0,0],[1,0,1]]
Output: 2
Explanation:
The cell (1, 1) is as far as possible from all the land with distance 2.
Example 2:
Input: [[1,0,0],[0,0,0],[0,0,0]]
Output: 4
Explanation:
The cell (2, 2) is as far as possible from all the land with distance 4.
"""
"""
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-1
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[0]) #求出地图的行列
        queue = [] #保存地图中出现'1'的位置
        for i in range(row):    #遍历地图。将为'1'的所有位置放入队列
            for j in range(col):
                if grid[i][j] == 1:
                    queue.append((i, j))
        if len(queue) == row*col or not queue: #如果地图全为'1'或者全为'0'，返回-1
            return -1
        level = 0   #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-1, y], [x+1, y], [x, y-1], [x, y+1]]: #每层扩张的四个方向
                    if 0 <= i <row and 0<= j < col and grid[i][j] == 0:
                        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] = 1    #并将搜索过的位置 变为1
            queue = newqueue
            level += 1
        return (level-1)