zoukankan      html  css  js  c++  java
  • Number of Islands——LeetCode

    Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. 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 1:

    11110
    11010
    11000
    00000

    Answer: 1

    Example 2:

    11000
    11000
    00100
    00011

    Answer: 3

    哈哈,这是今天的新题,十多分钟A了,题目不难,我的做法是BFS,题目大意是1代表岛屿,0代表水,可以假设这个grid周围都是水,岛屿定义是四周都是水,求岛屿的个数。

    我的做法是,从第一个开始,如果是1,并且没有访问过visit为false,那么加入queue里,当queue非空取出来并加入它上下左右的邻居,并且把这些节点visit设置为true,时间复杂度是O(M*N),空间复杂度的话,因为同时在queue里的最多也就4个节点的坐标,所以是O(1)。

    Talk is cheap>>

    public int numIslands(char[][] grid) {
            if (grid == null || grid.length == 0) {
                return 0;
            }
            int rowLen = grid.length;
            int colLen = grid[0].length;
            boolean[][] visited = new boolean[rowLen][colLen];
            int[][] direct = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
            Deque<Integer> queue = new ArrayDeque<>();
            int res = 0;
            for (int i = 0; i < rowLen; i++) {
                for (int j = 0; j < colLen; j++) {
                    if (!visited[i][j] && grid[i][j] == '1') {
                        res++;
                        queue.add(i * colLen + j);
                        while (!queue.isEmpty()) {
                            int pos = queue.poll();
                            int curr_x = pos / colLen;
                            int curr_y = pos % colLen;
                            if (visited[curr_x][curr_y]) {
                                continue;
                            }
                            visited[curr_x][curr_y] = true;
                            for (int k = 0; k < 4; k++) {
                                int x = curr_x + direct[k][0];
                                int y = curr_y + direct[k][1];
                                if (x >= 0 && y >= 0 && x < rowLen && y < colLen && grid[x][y] == '1') {
                                    queue.add(x * colLen + y);
                                }
                            }
                        }
                    }
                }
            }
            return res;
        }

  • 相关阅读:
    k8s中service 的iptables
    xbak 备份
    tcp/ip 拥塞控制、重传、丢包、优化
    mysql 主从详细原理 以及prom 监控的对象
    内核参数优化limit.conf与sysctl.conf
    k8s 中(生产|测试)环境隔离问题
    mysql锁以及配置优化
    mysql5.6迁移mysql5.7(生产中、短中断)
    mysql 日志方面与备份、恢复
    Apollo&&Eureka安装配置
  • 原文地址:https://www.cnblogs.com/aboutblank/p/4405557.html
Copyright © 2011-2022 走看看