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.
For 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
这道题类似一种迷宫问题，规定了-1表示墙，0表示门，让求每个点到门的最近的曼哈顿距离，这其实类似于求距离场Distance Map的问题，那么我们先考虑用DFS来解，思路是，我们搜索0的位置，每找到一个0，以其周围四个相邻点为起点，开始DFS遍历，并带入深度值1，如果遇到的值大于当前深度值，我们将位置值赋为当前深度值，并对当前点的四个相邻点开始DFS遍历，注意此时深度值需要加1，这样遍历完成后，所有的位置就被正确地更新了，参见代码如下：
解法一：
class Solution { public: void wallsAndGates(vector<vector<int>>& rooms) { for (int i = 0; i < rooms.size(); ++i) { for (int j = 0; j < rooms[i].size(); ++j) { if (rooms[i][j] == 0) { dfs(rooms, i + 1, j, 1); dfs(rooms, i - 1, j, 1); dfs(rooms, i, j + 1, 1); dfs(rooms, i, j - 1, 1); } } } } void dfs(vector<vector<int>> &rooms, int i, int j, int val) { if (i < 0 || i >= rooms.size() || j < 0 || j >= rooms[i].size()) return; if (rooms[i][j] > val) { rooms[i][j] = val; dfs(rooms, i + 1, j, val + 1); dfs(rooms, i - 1, j, val + 1); dfs(rooms, i, j + 1, val + 1); dfs(rooms, i, j - 1, val + 1); } } };
上述的代码可以稍稍简化一下，我们将rooms[i][j]和val的大小判断放入起始判断条件中，这样我们就可以以门的位置开始调用DFS函数了，参见代码如下：
解法二：
// DFS class Solution { public: void wallsAndGates(vector<vector<int>>& rooms) { for (int i = 0; i < rooms.size(); ++i) { for (int j = 0; j < rooms[i].size(); ++j) { if (rooms[i][j] == 0) { dfs(rooms, i, j, 0); } } } } void dfs(vector<vector<int>> &rooms, int i, int j, int val) { if (i < 0 || i >= rooms.size() || j < 0 || j >= rooms[i].size() || rooms[i][j] < val) return; rooms[i][j] = val; dfs(rooms, i + 1, j, val + 1); dfs(rooms, i - 1, j, val + 1); dfs(rooms, i, j + 1, val + 1); dfs(rooms, i, j - 1, val + 1); } };
那么下面我们再来看BFS的解法，却要借助queue，我们首先把门的位置都排入queue中，然后开始循环，对于门位置的四个相邻点，我们判断其是否在矩阵范围内，并且位置值是否大于上一位置的值加1，如果满足这些条件，我们将当前位置赋为上一位置加1，并将次位置排入queue中，这样等queue中的元素遍历完了，所有位置的值就被正确地更新了，参见代码如下：
解法三：
// BFS class Solution { public: void wallsAndGates(vector<vector<int>>& rooms) { queue<pair<int, int>> q; vector<vector<int>> dirs{{0, -1}, {-1, 0}, {0, 1}, {1, 0}}; for (int i = 0; i < rooms.size(); ++i) { for (int j = 0; j < rooms[i].size(); ++j) { if (rooms[i][j] == 0) q.push({i, j}); } } while (!q.empty()) { int i = q.front().first, j = q.front().second; q.pop(); for (int k = 0; k < dirs.size(); ++k) { int x = i + dirs[k][0], y = j + dirs[k][1]; if (x < 0 || x >= rooms.size() || y < 0 || y >= rooms[0].size() || rooms[x][y] < rooms[i][j] + 1) continue; rooms[x][y] = rooms[i][j] + 1; q.push({x, y}); } } } };
本文转自博客园Grandyang的博客，原文链接：墙和门[LeetCode] Walls and Gates ，如需转载请自行联系原博主。
