[LeetCode 286] Walls and Gates (Medium)

Solution

1. 从每个gate出发，对其上下左右的INF用DFS求其distance深度。
2. 起始distance == 0， 每进一次，distance + 1
3. base case 1： row & col Out of Bound
4. base case2: 保证是最小的distance，且避开gate 和 wall。

// ensure the min distance, and avoid wall and gate as well because distance is start from 1, wall value is -1, gate is 0
if (rooms [row][col] < distance)
    return;

代码

class Solution {
    public void wallsAndGates(int[][] rooms) {
        if (rooms == null || rooms.length == 0 || rooms [0].length == 0)
            return;
        
        for (int i = 0; i < rooms.length; i++) {
            for (int j = 0; j < rooms[0].length; j++) {
                if (rooms [i][j] == 0) {
                    wallsAndGatesHelper (rooms, i, j, 0);  //distance from gate to INF
                }
            }
        }
    }
    
    public void wallsAndGatesHelper (int[][] rooms, int row, int col, int distance) {
        // out of bound
        if (row < 0 || row >= rooms.length || col < 0 || col >= rooms[0].length)
            return;
        // ensure the min distance, and avoid wall and gate as well because distance is start from 1, wall value is -1, gate is 0
        if (rooms [row][col] < distance)
            return;
        
        int[][] directions = {{1, 0},{0, 1},{-1, 0},{0, -1}};
        rooms [row][col] = distance;
        
        for (int[] direction : directions) {
            wallsAndGatesHelper (rooms, row + direction [0], col + direction [1], distance + 1);
        }
    }
}