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
解题思路：
用DFS或者BFS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。
C++: 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 + 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);
}
}
};
C++: DFS II
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);
}
};
C++: 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});
}
}
}
};
