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，那么先考虑用 DFS 来解，思路是，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，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 = ; i < rooms.size(); ++i) {
for (int j = ; j < rooms[i].size(); ++j) {
if (rooms[i][j] == ) dfs(rooms, i, j, );
}
}
}
void dfs(vector<vector<int>>& rooms, int i, int j, int val) {
if (i < || i >= rooms.size() || j < || j >= rooms[i].size() || rooms[i][j] < val) return;
rooms[i][j] = val;
dfs(rooms, i + , j, val + );
dfs(rooms, i - , j, val + );
dfs(rooms, i, j + , val + );
dfs(rooms, i, j - , val + );
}
};
那么下面再来看 BFS 的解法，需要借助 queue，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，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，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：
解法二：
class Solution {
public:
void wallsAndGates(vector<vector<int>>& rooms) {
queue<pair<int, int>> q;
vector<vector<int>> dirs{{, -}, {-, }, {, }, {, }};
for (int i = ; i < rooms.size(); ++i) {
for (int j = ; j < rooms[i].size(); ++j) {
if (rooms[i][j] == ) q.push({i, j});
}
}
while (!q.empty()) {
int i = q.front().first, j = q.front().second; q.pop();
for (int k = ; k < dirs.size(); ++k) {
int x = i + dirs[k][], y = j + dirs[k][];
if (x < || x >= rooms.size() || y < || y >= rooms[].size() || rooms[x][y] < rooms[i][j] + ) continue;
rooms[x][y] = rooms[i][j] + ;
q.push({x, y});
}
}
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/286
类似题目：
Shortest Distance from All Buildings
Rotting Oranges
参考资料：
https://leetcode.com/problems/walls-and-gates/
https://leetcode.com/problems/walls-and-gates/discuss/72745/Java-BFS-Solution-O(mn)-Time