1 """ 2 Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell. 3 The distance between two adjacent cells is 1. 4 Example 1: 5 Input: 6 [[0,0,0], 7 [0,1,0], 8 [0,0,0]] 9 Output: 10 [[0,0,0], 11 [0,1,0], 12 [0,0,0]] 13 Example 2: 14 Input: 15 [[0,0,0], 16 [0,1,0], 17 [1,1,1]] 18 Output: 19 [[0,0,0], 20 [0,1,0], 21 [1,2,1]] 22 """ 23 """ 24 本题与leetcode1162题类似 25 先找出matrix中所有0的坐标,记录在队列q中， 26 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 27 遍历q中元素q[i]，更新q[i]周围四个点的距离， 28 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 29 """ 30 class Solution: 31 def updateMatrix(self, matrix): 32 queue = [] 33 m = len(matrix) # m行 34 n = len(matrix[0]) # n列 35 for x in range(m): 36 for y in range(n): 37 if matrix[x][y] == 0: # 记录0所在的位置 38 queue.append((x, y)) 39 else: 40 matrix[x][y] = m + n # 非0位置设为最大值 41 for x, y in queue: # 时间复杂度O(m*n*4) 42 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 43 if 0 <= i < m and 0 <= j < n and matrix[x][y] + 1 < matrix[i][j]: #注意边界0<=i<m 44 matrix[i][j] = matrix[x][y] + 1 #！！！记录距离 45 queue.append((i, j)) #将更改后距离的位置入队 46 return matrix