@@ -62,6 +62,93 @@ class Solution(object):
6262 return res
6363```
6464
65+ ## 542. 01 矩阵
66+ 67+ [ 原题链接] ( https://leetcode-cn.com/problems/01-matrix/ )
68+ 69+ ### BFS:以 1 为源(超时)
70+ 71+ ``` python
72+ class Solution :
73+ def updateMatrix (self matrix : List[List[int ]]) -> List[List[int ]]:
74+ m = len (matrix)
75+ if m == 0 :
76+ return []
77+ n = len (matrix[0 ])
78+ 79+ ans = [[0 for _ in range (n)] for _ in range (m)]
80+ 81+ for i in range (m):
82+ for j in range (n):
83+ if matrix[i][j] == 0 :
84+ # 0 不处理
85+ continue
86+ mark = [[0 for _ in range (n)] for _ in range (m)]
87+ step = 0
88+ queue = [(i, j, step)]
89+ while len (queue) > 0 :
90+ # bfs
91+ x, y, s = queue[0 ][0 ], queue[0 ][1 ], queue[0 ][2 ]
92+ del queue[0 ]
93+ if mark[x][y]:
94+ # 已经访问过,跳过
95+ continue
96+ # 处理
97+ mark[x][y] = 1 # 访问标记
98+ if matrix[x][y] == 0 :
99+ # 找到 0,进行标记,不继续遍历
100+ ans[i][j] = s
101+ break
102+ 103+ # 否则加入上下左右
104+ directions = [[0 , 1 ], [0 , - 1 ], [- 1 , 0 ], [1 , 0 ]]
105+ for d in directions:
106+ n_x, n_y = x + d[0 ], y + d[1 ]
107+ if n_x >= 0 and n_x < m and n_y >= 0 and n_y < n and mark[n_x][n_y] == 0 :
108+ # 坐标符合要求
109+ # print(n_x, n_y, s + 1)
110+ queue.append((n_x, n_y, s + 1 ))
111+ 112+ return ans
113+ ```
114+ 115+ ### BFS:以 0 为源
116+ 117+ 把所有 0 放入队列,每个 0 逐个向外 BFS 一圈标记相邻的 1,再把 BFS 到的 1 入队列......
118+ 119+ ``` python
120+ class Solution :
121+ def updateMatrix (self matrix : List[List[int ]]) -> List[List[int ]]:
122+ m = len (matrix)
123+ if m == 0 :
124+ return []
125+ n = len (matrix[0 ])
126+ 127+ queue = []
128+ 129+ for i in range (m):
130+ for j in range (n):
131+ # 把 0 放入队列
132+ if matrix[i][j] == 0 :
133+ queue.append((i, j))
134+ else :
135+ # 把 1 标记为未访问过的结点
136+ matrix[i][j] = - 1
137+ directions = [[0 , 1 ], [0 , - 1 ], [1 , 0 ], [- 1 , 0 ]]
138+ while len (queue) > 0 :
139+ x, y = queue[0 ][0 ], queue[0 ][1 ]
140+ del queue[0 ]
141+ for d in directions:
142+ n_x, n_y = x + d[0 ], y + d[1 ]
143+ if n_x >= 0 and n_x < m and n_y >= 0 and n_y < n and matrix[n_x][n_y] == - 1 :
144+ matrix[n_x][n_y] = matrix[x][y] + 1
145+ queue.append((n_x, n_y))
146+ 147+ return matrix
148+ ```
149+ 150+ - 时间复杂度:$O(m * n)$
151+ 65152## 1162. 地图分析
66153
67154[ 原题链接] ( https://leetcode-cn.com/problems/as-far-from-land-as-possible/ )
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