@@ -79,6 +79,55 @@ def BFS(x):
7979
8080## 例题
8181
82+ ### [ walls-and-gates] ( https://leetcode-cn.com/problems/walls-and-gates/ )
83+ 84+ > 给定一个二维矩阵,矩阵中元素 -1 表示墙或是障碍物,0 表示一扇门,INF (2147483647) 表示一个空的房间。你要给每个空房间位上填上该房间到最近门的距离,如果无法到达门,则填 INF 即可。
85+
86+ ** 图森面试真题** 。典型的多源最短路径问题,将所有源作为 BFS 的第一层即可
87+ 88+ ``` Python
89+ inf = 2147483647
90+ 91+ class Solution :
92+ def wallsAndGates (self rooms : List[List[int ]]) -> None :
93+ 94+ if not rooms or not rooms[0 ]:
95+ return
96+ 97+ M, N = len (rooms), len (rooms[0 ])
98+ 99+ bfs = collections.deque([])
100+ 101+ for i in range (M):
102+ for j in range (N):
103+ if rooms[i][j] == 0 :
104+ rooms[i][j] = inf
105+ bfs.append((i, j))
106+ 107+ dist = 0
108+ while bfs:
109+ num_level = len (bfs)
110+ for _ in range (num_level):
111+ r, c = bfs.popleft()
112+ if rooms[r][c] == inf:
113+ rooms[r][c] = dist
114+ 115+ if r - 1 >= 0 and rooms[r - 1 ][c] == inf:
116+ bfs.append((r - 1 , c))
117+ 118+ if r + 1 < M and rooms[r + 1 ][c] == inf:
119+ bfs.append((r + 1 , c))
120+ 121+ if c - 1 >= 0 and rooms[r][c - 1 ] == inf:
122+ bfs.append((r, c - 1 ))
123+ 124+ if c + 1 < N and rooms[r][c + 1 ] == inf:
125+ bfs.append((r, c + 1 ))
126+ dist += 1
127+ 128+ return
129+ ```
130+ 82131### [ shortest-bridge] ( https://leetcode-cn.com/problems/shortest-bridge/ )
83132
84133> 在给定的 01 矩阵 A 中,存在两座岛 (岛是由四面相连的 1 形成的一个连通分量)。现在,我们可以将 0 变为 1,以使两座岛连接起来,变成一座岛。返回必须翻转的 0 的最小数目。
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