286. Walls and Gates

 

You are given a m x n 2D grid initialized with these three possible values.

  1. -1 - A wall or an obstacle.
  2. 0 - A gate.
  3. INF - Infinity means an empty room. We use the value 2 31 - 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.

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 = 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{{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});
            }
        }
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/286

 

类似题目:

Surrounded Regions

Number of Islands

Shortest Distance from All Buildings

Robot Room Cleaner

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

https://leetcode.com/problems/walls-and-gates/discuss/72746/My-short-java-solution-very-easy-to-understand

  

LeetCode All in One 题目讲解汇总(持续更新中…)


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