[LintCode] 598 Zombie in Matrix 解题报告

Description
Given a 2D grid, each cell is either a wall 2, a zombie 1 or people 0 (the number zero, one, two).Zombies can turn the nearest people(up/down/left/right) into zombies every day, but can not through wall. How long will it take to turn all people into zombies? Return -1 if can not turn all people into zombies.

Example
Given a matrix:

0 1 2 0 0
1 0 0 2 1
0 1 0 0 0
return 2

5/13/2017

算法班，未经验证

注意，这里需要算需要多久，所以是层序遍历的问题，所以第39行开始是层序遍历的写法，普通BFS不需要，而且在放入queue之前就要先判断好是否是1，否则会无谓的监测一遍，day的最后值有出入

 1 public class Solution {
 2     /**
 3      * @param grid  a 2D integer grid
 4      * @return an integer
 5      */
 6
 7     class Point{
 8         int x;
 9         int y;
10         public Point(int x, int y) {
11             this.x = x;
12             this.y = y;
13         }
14     }
15     public int zombie(int[][] grid) {
16
17         // Write your code here
18         if (grid == null || grid.length == 0 || grid[0].length == 0) {
19             return -1;
20         }
21         int[] xDirection = new int[]{1, -1, 0, 0};
22         int[] yDirection = new int[]{0, 0, 1, -1};
23
24         // boolean[][] visited = new boolean[grid.length][grid[0].length];
25         Queue<Point> queue = new LinkedList<Point>();
26         int people = 0, day = 0;
27
28         for (int i = 0; i < grid.length; i++) {
29             for (int j = 0; j < grid[0].length; j++) {
30                 if (grid[i][j] == 1) {
31                     queue.offer(new Point(i, j));
32                 } else if (grid[i][j] == 0) {
33                     people++;
34                 }
35             }
36         }
37
38         while (!queue.isEmpty()) {
39             int size = queue.size();
40             for (int i = 0; i < size; i++) {
41                 Poing p = queue.poll();
42                 for (int j = 0; j < xDirection.length; i++) {
43                     int tmpX = p.x + xDirection[j];
44                     int tmpY = p.y + yDirection[j];
45                     if (inBoundary(tmpX, tmpY, grid.length, grid[0].length) && grid[tmpX][tmpY] == 0) {
46                         queue.offer(new Point(tmpX, tmpY));
47                         grid[tmpX][tmpY] = 1;
48                         people--;
49                     }
50                 }
51             }
52             day++;
53             if (people == 0) {
54                 return day;
55             }
56         }
57         return -1;
58     }
59
60     private boolean inBoundary(int x, int y, int xLength, int yLength) {
61         if (x < xLength && x >= 0 && y < yLength && y >= 0) return true;
62         return false;
63     }
64
65 }