leetcode542 01 Matrix

"""
Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.
The distance between two adjacent cells is 1.
Example 1:
Input:
[[0,0,0],
 [0,1,0],
 [0,0,0]]
Output:
[[0,0,0],
 [0,1,0],
 [0,0,0]]
Example 2:
Input:
[[0,0,0],
 [0,1,0],
 [1,1,1]]
Output:
[[0,0,0],
 [0,1,0],
 [1,2,1]]
"""
"""
本题与leetcode1162题类似
先找出matrix中所有0的坐标,记录在队列q中，
并将值为1的地方赋为n+m，因为最远的距离也不会超过n+m
遍历q中元素q[i]，更新q[i]周围四个点的距离，
更新了距离的点的坐标入队，直到遍历到q的最后一个元素为止
"""
class Solution:
    def updateMatrix(self, matrix):
        queue = []
        m = len(matrix)  # m行
        n = len(matrix[0]) # n列
        for x in range(m):
            for y in range(n):
                if matrix[x][y] == 0:    # 记录0所在的位置
                    queue.append((x, y))
                else:
                    matrix[x][y] = m + n  # 非0位置设为最大值
        for x, y in queue: # 时间复杂度O(m*n*4)
            for i, j in [(x+1, y), (x-1, y), (x, y+1), (x, y-1)]: #！！！上下左右四个方向遍历
                if 0 <= i < m and 0 <= j < n and matrix[x][y] + 1 < matrix[i][j]: #注意边界0<=i<m
                    matrix[i][j] = matrix[x][y] + 1 #！！！记录距离
                    queue.append((i, j)) #将更改后距离的位置入队
        return matrix