LeetCode 题解(257) : Walls and Gates_leetcode c语言 wallsandgates-优快云博客

本文介绍了一种用于求解二维网格中空房间到门最短距离的问题，通过遍历算法更新每个空房间到最近门的距离，并提供了一种使用队列优化的实现方式。

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题目：

ou 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.

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

题解：

先自己写了一下，注意优化条件是相邻元素的值如果大于当前值加1，就不需考虑相邻元素了：

class Solution {
public:
    void wallsAndGates(vector<vector<int>>& rooms) {
        if(rooms.size() == 0)
            return;
        vector<vector<bool>> visited(rooms.size(), vector<bool>(rooms[0].size(), false));
        for(int i = 0; i < rooms.size(); i++) {
            for(int j = 0; j < rooms[0].size(); j++) {
                if(rooms[i][j] == 0) {
                    if (i - 1 >= 0 && rooms[i - 1][j] > rooms[i][j] + 1)    
                        distance(rooms, visited, i - 1, j, 1);
                    if (i + 1 < rooms.size() && rooms[i + 1][j] > rooms[i][j] + 1)
                        distance(rooms, visited, i + 1, j, 1);
                    if (j - 1 >= 0 && rooms[i][j - 1] > rooms[i][j] + 1)
                        distance(rooms, visited, i, j - 1, 1);
                    if (j + 1 < rooms[0].size() && rooms[i][j + 1] > rooms[i][j] + 1)
                        distance(rooms, visited, i, j + 1, 1);
                }
            }
        }
    }
    
    void distance(vector<vector<int>>& rooms, vector<vector<bool>>& visited, int i, int j, int d) {
        if (i < 0 || i > rooms.size() - 1 || j < 0 || j > rooms[0].size() - 1)
            return;
        if (rooms[i][j] == -1 || rooms[i][j] == 0)
            return;
        if (visited[i][j])
            rooms[i][j] = min(rooms[i][j], d);
        else {
            rooms[i][j] = d;
            visited[i][j] = true;
        }
        if (i - 1 >= 0 && rooms[i - 1][j] > rooms[i][j] + 1)    
            distance(rooms, visited, i - 1, j, d + 1);
        if (i + 1 < rooms.size() && rooms[i + 1][j] > rooms[i][j] + 1)
            distance(rooms, visited, i + 1, j, d + 1);
        if (j - 1 >= 0 && rooms[i][j - 1] > rooms[i][j] + 1)
            distance(rooms, visited, i, j - 1, d + 1);
        if (j + 1 < rooms[0].size() && rooms[i][j + 1] > rooms[i][j] + 1)
            distance(rooms, visited, i, j + 1, d + 1);
    }
};

网上学来Queue的简便写法：