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  • 286. Walls and Gates (Solution 2)

    package LeetCode_286
    
    import java.util.*
    
    /**
     * 286. Walls and Gates
     * (Prime)
     *
     * 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 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.
    
    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
     * */
    class Solution {
        /*
        * Solution 1: DFS, Time complexity:O(m^2*n^2), Space complexity:O(4^n)
        * Solution 2: BFS, Time complexity:O(mn), Space complexity:O(mn)
        * */
        fun fillEmptyRoom(grid: Array<IntArray>) {
            val m = grid.size
            val n = grid[0].size
            for (i in 0 until m) {
                for (j in 0 until n) {
                    //start bfs from where is a gate
                    if (grid[i][j] == 0) {
                        bfs(Pair(i, j), grid)
                    }
                }
            }
       }
    
        private fun bfs(pair: Pair<Int, Int>, grid: Array<IntArray>) {
            val directions = intArrayOf(0, 1, 0, -1, 0)
            val queue = LinkedList<Pair<Int, Int>>()
            queue.offer(pair)
            while (queue.isNotEmpty()) {
                val cur = queue.pop()
                //expand 4 directions
                for (i in 0 until 4) {
                    val x = cur.first + directions[i]
                    val y = cur.second + directions[i + 1]
                    //return when reach gate,wall,or confirm min distance
                    if (x < 0 || x >= grid.size || y < 0 || y >= grid[0].size || grid[x][y] < grid[cur.first][cur.second] + 1) {
                        continue
                    }
                    //update new count
                    grid[x][y] = grid[cur.first][cur.second] + 1
                    queue.offer(Pair(x, y))
                }
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13553302.html
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