[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

public class Solution {
    /**
     * @param grid  a 2D integer grid
     * @return an integer
     */
     
    class Point{
        int x;
        int y;
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    } 
    public int zombie(int[][] grid) {

        // Write your code here
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return -1;
        }
        int[] xDirection = new int[]{1, -1, 0, 0};
        int[] yDirection = new int[]{0, 0, 1, -1};

        // boolean[][] visited = new boolean[grid.length][grid[0].length];
        Queue<Point> queue = new LinkedList<Point>();
        int people = 0, day = 0;

        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    queue.offer(new Point(i, j));
                } else if (grid[i][j] == 0) {
                    people++;
                }
            }
        }

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                Poing p = queue.poll();
                for (int j = 0; j < xDirection.length; i++) {
                    int tmpX = p.x + xDirection[j];
                    int tmpY = p.y + yDirection[j];
                    if (inBoundary(tmpX, tmpY, grid.length, grid[0].length) && grid[tmpX][tmpY] == 0) {
                        queue.offer(new Point(tmpX, tmpY));
                        grid[tmpX][tmpY] = 1;
                        people--;
                    }
                }
            }
            day++;
            if (people == 0) {
                return day;
            }
        }
        return -1;
    }

    private boolean inBoundary(int x, int y, int xLength, int yLength) {
        if (x < xLength && x >= 0 && y < yLength && y >= 0) return true;
        return false;
    }

}