A 2d grid map of m rows and n columns is initially filled with water. We may perform an addLand operation which turns the water at position (row, col) into a land. Given a list of positions to operate, count the number of islands after each addLand operation. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example:

Given m = 3, n = 3positions = [[0,0], [0,1], [1,2], [2,1]].
Initially, the 2d grid grid is filled with water. (Assume 0 represents water and 1 represents land).

0 0 0
0 0 0
0 0 0

Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land.

1 0 0
0 0 0 Number of islands = 1
0 0 0

Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land.

1 1 0
0 0 0 Number of islands = 1
0 0 0

Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land.

1 1 0
0 0 1 Number of islands = 2
0 0 0

Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land.

1 1 0
0 0 1 Number of islands = 3
0 1 0

We return the result as an array: [1, 1, 2, 3]

public List<Integer> numIslands2(int m, int n, int[][] positions) {
int[] rootArray = new int[m*n];
Arrays.fill(rootArray,-1); ArrayList<Integer> result = new ArrayList<Integer>(); int[][] directions = {{-1,0},{0,1},{1,0},{0,-1}};
int count=0; for(int k=0; k<positions.length; k++){
count++; int[] p = positions[k];
int index = p[0]*n+p[1];
rootArray[index]=index;//set root to be itself for each node for(int r=0;r<4;r++){
int i=p[0]+directions[r][0];
int j=p[1]+directions[r][1]; if(i>=0&&j>=0&&i<m&&j<n&&rootArray[i*n+j]!=-1){
//get neighbor's root
int thisRoot = getRoot(rootArray, i*n+j);
if(thisRoot!=index){
rootArray[thisRoot]=index;//set previous root's root
count--;
}
}
} result.add(count);
} return result;
} public int getRoot(int[] arr, int i){
while(i!=arr[i]){
i=arr[i];
}
return i;
}

二刷:

public class Solution {
public List<Integer> numIslands2(int m, int n, int[][] positions) {
int[] id = new int[m * n]; // 表示各个index对应的root List<Integer> res = new ArrayList<>();
Arrays.fill(id, -1); // 初始化root为-1,用来标记water, 非-1表示land
int count = 0; // 记录island的数量 int[][] dirs = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
for (int i = 0; i < positions.length; i++) {
count++;
int index = positions[i][0] * n + positions[i][1];
id[index] = index; // root初始化 for (int j = 0; j < dirs.length; j++) {
int x = positions[i][0] + dirs[j][0];
int y = positions[i][1] + dirs[j][1];
if (x >= 0 && x < m && y >= 0 && y < n && id[x * n + y] != -1) {
int root = root(id, x * n + y); // 发现root不等的情况下,才union, 同时减小count
if (root != index) {
id[root] = index;
count--;
}
}
}
res.add(count);
}
return res;
} public int root(int[] id, int i) {
while (i != id[i]) {
id[i] = id[id[i]]; // 优化,为了减小树的高度
i = id[i];
}
return i;
}
}

最新文章

  1. ES6 箭头函数中的 this?你可能想多了(翻译)
  2. [转]Unicode utf8等编码类型的原理
  3. 解决git无法clone提示443以及配置git代理方法
  4. EasyUI中Treegrid节点的删除
  5. hbase shell中log4j重复问题
  6. 阿里云安装nginx 和 php-fpm
  7. [未完成][Mooc]关于Linxu的总结(一)
  8. HTML&amp;CSS基础学习笔记1.30-颜色的表达
  9. Java 7 Fork/Join 框架
  10. git fetch, git pull 剖析
  11. Linux新手随手笔记1.2
  12. web Deploy发布问题
  13. EasyUI 学习(1)-Tooltip(提示框)
  14. MySQL数据库的基础学习
  15. PHP性能监测的工具介绍 - XHProf -参考自https://jingyan.baidu.com/article/7082dc1c173359e40a89bd95.html
  16. 【Hibernate步步为营】--最后的集合映射
  17. 1047 行 MySQL 详细学习笔记
  18. thinkphp 如何调用百度echarts 数据报表插件
  19. objectARX加载lisp函数、源码的一种方式
  20. 【费用流】BZOJ1877[SDOI2009]-晨跑

热门文章

  1. Windows Server 2016与旧版本系统比较
  2. images
  3. bootstarp3
  4. 7.5 C++基本序列式容器
  5. 1.5 socket服务器传输文件
  6. Convert the AScii to SAC file
  7. format格式
  8. HTTP网页过程
  9. GIL 相关 和进程池
  10. git使用简明教程