- 作者: 负雪明烛
- id: fuxuemingzhu
- 个人博客:http://fuxuemingzhu.cn/
目录
- BFS
- 日期
题目地址:https://leetcode-cn.com/problems/walls-and-gates/
题目描述
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 value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.
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
题目大意
你被给定一个 m × n 的二维网格,网格中有以下三种可能的初始化值:
- -1 表示墙或是障碍物
- 0 表示一扇门
- INF 无限表示一个空的房间。然后,我们用 231 - 1 = 2147483647 代表 INF。你可以认为通往门的距离总是小于 2147483647 的。
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。
解题方法
BFS
这个题可以使用BFS和DFS去解决,我选用的BFS。
- 找出所有为0的位置,放入队列中
- 从为0的位置开始向四个方向遍历,直接修改迷宫的可以走的位置的数字是距离门最近的距离
- 把新的可以走的位置放入队列中,继续搜索
C++代码如下:
class Solution {
public:
void wallsAndGates(vector<vector<int>>& rooms) {
if (rooms.empty() || rooms[0].empty()) return;
M = rooms.size();
N = rooms[0].size();
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (rooms[i][j] == 0) {
bfs(rooms, i, j);
}
}
}
}
void bfs(vector<vector<int>>& rooms, int x, int y) {
if (rooms[x][y] != 0) return;
queue<vector<int>> que;
que.push({x, y});
while (!que.empty()) {
vector<int> cur = que.front(); que.pop();
for (auto& dir : dirs) {
int newx = cur[0] + dir[0];
int newy = cur[1] + dir[1];
if (newx < 0 || newx >= M || newy < 0 || newy >= N
|| rooms[newx][newy] <= rooms[cur[0]][cur[1]])
continue;
rooms[newx][newy] = rooms[cur[0]][cur[1]] + 1;
que.push({newx, newy});
}
}
}
private:
int M, N;
vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
};
日期
2019 年 9 月 23 日 —— 昨夜睡的早,错过了北京的烟火