[LeetCode] 286. Walls and Gates_Medium tag: BFS
2018-07-06 06:19 Johnson_强生仔仔 阅读(243) 评论(0) 编辑 收藏 举报You are given a m x n 2D grid initialized with these three possible values.
-1- A wall or an obstacle.0- A gate.INF- Infinity means an empty room. We use the value231 - 1 = 2147483647to representINFas you may assume that the distance to a gate is less than2147483647.
Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.
Example:
Given the 2D grid:
INF -1 0 INF INF INF INF -1 INF -1 INF -1 0 -1 INF INF
After running your function, the 2D grid should be:
3 -1 0 1 2 2 1 -1 1 -1 2 -1 0 -1 3 4
这个题目是用BFS来解决, 最naive approach就是扫每个点,然后针对每个点用BFS一旦找到0, 代替点的值. 时间复杂度为 O((m*n)^2);
我们还是用BFS的思路, 但是我们先把2D arr 扫一遍, 然后把是0的位置都append进入queue里面, 因为是BFS, 所以每一层最靠近0的位置都会被替换, 并且标记为visited, 然后一直BFS到最后即可.
时间复杂度降为O(m*n)
1. Constraints
1) empty or len(rooms[0]) == 0, edge case
2) size of the rooms < inf
3) 每个元素的值为0, -1, inf
2. Ideas
BFS T: O(m*n) S: O(m*n)
1) edge case,empty or len(rooms[0]) == 0 => return
2) scan 2D arr, 将值为0 的append进入queue里面
3) BFS, 如果是not visited过得, 并且 != 0 or -1, 更改值为dis + 1, 另外append进入queue, add进入visited
3. Code
1 class Solution: 2 def wallAndGates(self, rooms): 3 if not rooms or len(rooms[0]) == 0: return 4 queue, lr, lc, visited, dirs = collections.deque(), len(rooms), len(rooms[0]), set(), [(1, 0), (-1, 0), (0, 1), (0, -1)] 5 for i in range(lr): 6 for j in range(lc): 7 if rooms[i][j] == 0: 8 queue.append((i, j, 0)) 9 visited.add((i, j))
10 while queue: 11 pr, pc, dis = queue.popleft() 12 for d1, d2 in dirs: 13 nr, nc = pr + d1, pc + d2 14 if 0<= nr < lr and 0<= nc < lc and (nr, nc) not in visited and rooms[nr][nc] != -1: 15 rooms[nr][nc] = dis + 1 16 queue.append((nr,nc, dis + 1)) 17 visited.add((nr, nc))
4. Test cases
1) edge case
2)
INF -1 0 INF INF INF INF -1 INF -1 INF -1 0 -1 INF INF
After running function, the 2D grid should be:
3 -1 0 1 2 2 1 -1 1 -1 2 -1 0 -1 3 4
