286. Walls and Gates (Solution 2)
阿新 • • 發佈:2020-08-24
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)) } } } }