标签：ntc day zomb this new 矩阵 tco lis number
Given a 2D grid, each cell is either a wall 2, a zombie 1or 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.
Given a matrix:
0 1 2 0 0
1 0 0 2 1
0 1 0 0 0
return 2
1. BFS。queue+size层级遍历。第一层是初始的僵尸。每天晚上推入僵尸旁边的新僵尸（注意处理过的僵尸不用再次处理了，因为不会再有影响，靠周围那圈才是新生力量）。遍历的同时用一个count记录剩余人类。人减到0终止遍历，或者queue空了终止遍历（僵尸不够格）。
2.for循环。子函数每天做僵尸的咬人转换。主函数每天找到所有僵尸都做一次transfer。这个比较简单，但会有重复僵尸，效率不如前面的高。
细节：1.变量定义，最前面最好写下int ZOMBIE = 1; 这种，更直观。2.这种矩阵的都小心索引边界确认+dxdy简洁写法。3.二维数组调用还是统一下吧，函数头写int x, int y，矩阵调用也用int[x][y]，x和h比，y和w比。相当于把x看成向下指的坐标轴，y看成向右指的坐标轴，这样也自洽的。4. dxdy写法规范直接int[] dx = {-1, 0, 1, 0}; 等号右边不用写int说明
1.BFS：
public class Solution { /* * @param grid: a 2D integer grid * @return: an integer */ private 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 h = grid.length; int w = grid[0].length; int HUMAN = 0; int ZOMBIE = 1; int WALL = 2; int[] dx = {-1, 0, 1, 0}; int[] dy = {0, -1, 0, 1}; int humanCnt = 0; int dayCnt = 0; Queue<Point> queue = new LinkedList<Point>(); for (int i = 0; i < h; i++) { for (int j = 0; j < w; j++) { if (grid[i][j] == HUMAN) { humanCnt++; } else if (grid[i][j] == ZOMBIE) { queue.offer(new Point(i,j)); } } } if (humanCnt == 0) { return 0; } while (!queue.isEmpty()) { dayCnt++; // 你一开始忘了层级遍历！！！傻了把，那你就是每一个僵尸咬一次就加一天了 int size = queue.size(); for(int c = 0; c < size; c++) { Point crt = queue.poll(); for (int i = 0; i < 4; i++) { int cx = crt.x + dx[i]; int cy = crt.y + dy[i]; if (isValid(grid, cx, cy) && grid[cx][cy] == HUMAN) { grid[cx][cy] = ZOMBIE; queue.offer(new Point(cx,cy)); humanCnt--; if (humanCnt == 0) { return dayCnt; } } } } } return -1; } private boolean isValid(int[][] grid, int x, int y) { int h = grid.length; int w = grid[0].length; return x >= 0 && x < h && y >= 0 && y < w; } }
2.for循环
public class Solution { /* * @param grid: a 2D integer grid * @return: an integer */ public int zombie(int[][] grid) { // write your code here if (grid == null || grid.length == 0 || grid[0].length == 0) { return -1; } int h = grid.length; int w = grid[0].length; boolean hasChange = true; int dayCount =0; while (hasChange) { dayCount++; hasChange = false; for (int i = 0; i < h; i++) { for(int j = 0; j < w; j++) { if (grid[i][j] == 1 && transfer(grid, i, j)) { hasChange = true; } } } for (int i = 0; i < h; i++) { for (int j = 0; j < w; j++) { if (grid[i][j] == 3) { grid[i][j] = 1; } } } } for (int i = 0; i < h; i++) { for (int j = 0; j < w; j++) { if (grid[i][j] == 0) { return -1; } } } return dayCount - 1; } private boolean transfer(int[][] grid, int x, int y) { int[] dx = {-1, 0, 1, 0}; int[] dy = {0, -1, 0, 1}; boolean hasChange = false; for (int i = 0; i < 4; i++) { int cx = x + dx[i]; int cy = y + dy[i]; if (isValid(grid, cx, cy) && grid[cx][cy] == 0) { grid[cx][cy] = 3; hasChange = true; } } return hasChange; } private boolean isValid(int[][] grid, int x, int y) { int h = grid.length; int w = grid[0].length; return x >= 0 && x < h && y >= 0 && y < w; } }
lintcode598- Zombie in Matrix- medium
标签：ntc day zomb this new 矩阵 tco lis number
原文地址：http://www.cnblogs.com/jasminemzy/p/7745905.html
