zoukankan      html  css  js  c++  java
  • Leetcode: 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.
    
    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

    Push all gates into queue first. Then for each gate update its neighbor cells who is empty and push them to the queue.

     Repeating above steps until there is nothing left in the queue.

     1 public class Solution {
     2     int[][] dirs = new int[][] {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
     3     
     4     public void wallsAndGates(int[][] rooms) {
     5         if (rooms.length == 0 || rooms[0].length == 0) return;
     6         Queue<int[]> queue = new LinkedList<>();
     7         for (int i = 0; i < rooms.length; i++) {
     8             for (int j = 0; j < rooms[0].length; j++) {
     9                 if (rooms[i][j] == 0) queue.add(new int[]{i, j});
    10             }
    11         }
    12         int steps = 0;
    13         while (!queue.isEmpty()) {
    14             int size = queue.size();
    15             for (int i = 0; i < size; i ++) {
    16                 int[] cur = queue.poll();
    17                 for (int[] dir : dirs) {
    18                     int x = cur[0] + dir[0];
    19                     int y = cur[1] + dir[1];
    20                     if (x >= 0 && x < rooms.length && y >= 0 && y < rooms[0].length && rooms[x][y] == Integer.MAX_VALUE) {
    21                         rooms[x][y] = steps + 1;
    22                         queue.offer(new int[]{x, y});
    23                     }
    24                 }
    25             }
    26             steps ++;
    27         }
    28     }
    29 }

    Another way: similar

     1 public class Solution {
     2     public void wallsAndGates(int[][] rooms) {
     3         if (rooms.length == 0 || rooms[0].length == 0) return;
     4         Queue<int[]> queue = new LinkedList<>();
     5         for (int i = 0; i < rooms.length; i++) {
     6             for (int j = 0; j < rooms[0].length; j++) {
     7                 if (rooms[i][j] == 0) queue.add(new int[]{i, j});
     8             }
     9         }
    10         while (!queue.isEmpty()) {
    11             int[] top = queue.remove();
    12             int row = top[0], col = top[1];
    13             if (row > 0 && rooms[row - 1][col] == Integer.MAX_VALUE) {
    14                 rooms[row - 1][col] = rooms[row][col] + 1;
    15                 queue.add(new int[]{row - 1, col});
    16             }
    17             if (row < rooms.length - 1 && rooms[row + 1][col] == Integer.MAX_VALUE) {
    18                 rooms[row + 1][col] = rooms[row][col] + 1;
    19                 queue.add(new int[]{row + 1, col});
    20             }
    21             if (col > 0 && rooms[row][col - 1] == Integer.MAX_VALUE) {
    22                 rooms[row][col - 1] = rooms[row][col] + 1;
    23                 queue.add(new int[]{row, col - 1});
    24             }
    25             if (col < rooms[0].length - 1 && rooms[row][col + 1] == Integer.MAX_VALUE) {
    26                 rooms[row][col + 1] = rooms[row][col] + 1;
    27                 queue.add(new int[]{row, col + 1});
    28             }
    29         }
    30     }
    31 }
  • 相关阅读:
    前端日常开发常用功能系列之乱序
    ES6系列之箭头函数
    ES6系列之let/const及块级作用域
    前端日常开发常用功能系列之数组扁平
    前端日常开发常用功能系列之数组最值
    前端日常开发常用功能系列之拷贝
    前端日常开发常用功能系列之数组去重
    前端日常开发常用功能系列之节流
    前端日常开发常用功能系列之防抖
    数据库连接池优化配置(druid,dbcp,c3p0)
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5077950.html
Copyright © 2011-2022 走看看