[LeetCode] Walls and Gates

最新推荐文章于 2025-09-11 18:44:42 发布

转载最新推荐文章于 2025-09-11 18:44:42 发布 · 46 阅读

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原文链接：https://segmentfault.com/a/1190000009774807

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#数据结构与算法

本文介绍了一种使用广度优先搜索(BFS)算法来解决迷宫中每个空房间到最近出口的距离问题的方法。该算法首先将所有出口坐标加入队列，并从这些点开始遍历周围的空房间，更新其到出口的距离。通过这种方式逐步扩展直至所有可达区域都被标记。

Walls and Gates

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 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.

BFS

Time Complexity
O(MN)
Space Complexity
O(N)

思路

Push all gates into queue first. Then for each gate update its neighbor cells and push them to the queue. For visited, I just check if the matrixi for 4 directions is bigger than the current matrixi, if it is bigger means it hasn't been visited.

代码

public void wallsAndGates(int[][] rooms) {
    //corner case
    if(rooms == null || rooms.length == 0 || rooms[0] == null || rooms[0].length == 0) return;
    Queue<int[]> queue = new LinkedList<int[]>();
    int step = 1;
    int rows = rooms.length, cols = rooms[0].length;
    for(int i = 0; i < rows; i++){
        for(int j = 0; j < cols; j++){
            if(rooms[i][j] == 0) queue.offer(new int[]{i, j});
        }
    }
    
    int[][] directions = new int[][]{{-1, 0},{1, 0},{0,1},{0,-1}};
    
    while(!queue.isEmpty()){
        int size = queue.size();
        for(int i = 0; i < size; i++){
            int[] cur = queue.poll();
            for(int[] dir : directions){
                int x = cur[0] + dir[0];
                int y = cur[1] + dir[1];
                if(x >= 0 && x < rows && y >= 0 && y < cols && rooms[x][y] > rooms[cur[0]][cur[1]]){
                    rooms[x][y] = step;
                    queue.offer(new int[]{x,y});
                }
            }
        }
        step++;
    }
}