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Zombie in Matrix Lintcode
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
In this question, I store i * m + j in the queue, and m is the column number in the matrix.
In this question, pay attention that I should count the day first, becasue in the last day, everyone is zombie but if I don‘t return the day, the queue will keep iterate because it is not empty. For example, if I return the day number in the first day, the day should be 1. So add the day at first.
And pay attention that I can count the people first so if it drop to 0 I can return directly.
public class Solution { /** * @param grid a 2D integer grid * @return an integer */ public int zombie(int[][] grid) { if (grid == null || grid.length == 0 || grid[0].length == 0) { return -1; } int n = grid.length; int m = grid[0].length; Queue<Integer> q = new LinkedList<>(); int people = 0; for (int i = 0; i < grid.length; i++) { for (int j = 0; j < grid[0].length; j++) { if (grid[i][j] == 1) { q.offer(i * m + j); } if (grid[i][j] == 0) { people++; } } } int[] dx = {0, 0, 1, -1}; int[] dy = {1, -1, 0, 0}; int day = 0; while (!q.isEmpty()) { day++; int size = q.size(); for (int i = 0; i < size; i++) { int point = q.poll(); int x = point / m; int y = point % m; for (int j = 0; j < 4; j++) { int nx = x + dx[j]; int ny = y + dy[j]; if (isValid(grid, nx, ny) && grid[nx][ny] == 0) { q.offer(nx * m + ny); grid[nx][ny] = 1; people--; if (people == 0) { return day; } } } } } return -1; } private boolean isValid(int[][] grid, int x, int y) { return x >= 0 && y >= 0 && x < grid.length && y < grid[0].length; } }
Zombie in Matrix Lintcode
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