package leetcode_1To300;

import java.util.LinkedList;
import java.util.Queue;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _200_NumberofIslands {

    /**
     * 200. Number of Islands
     * 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

     time : O(m * n)
     space : O(n)

     * @param grid
     * @return
     */

    private int m;
    private int n;

    public int numIslands(char[][] grid) {
        int res = 0;
        m = grid.length;
        if (m == 0) return 0;
        n = grid[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    dfs(grid, i, j);
                    res++;
                }
            }
        }
        return res;
    }

    /**
     time : O(m * n)
     space : O(m * n)
     * @param grid
     * @param i
     * @param j
     */

    private void dfs(char[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= m || j >= n || grid[i][j] == '0') return;
        grid[i][j] = '0';
        dfs(grid, i, j + 1);
        dfs(grid, i, j - 1);
        dfs(grid, i + 1, j);
        dfs(grid, i - 1, j);
    }

    public int numIslands2(char[][] grid) {
        int res = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; i++) {
                if (grid[i][j] == '1') {
                    bfs(grid, i, j);
                    res++;
                }
            }
        }
        return res;
    }

    private void bfs(char[][] grid, int x, int y) {
        grid[x][y] = '0';
        int n = grid.length;
        int m = grid[0].length;
        Queue<Integer> queue = new LinkedList<>();
        int code = x * m + y;
        queue.offer(code);
        while (!queue.isEmpty()) {
            code = queue.poll();
            int i = code / m;
            int j = code % m;
            if (i > 0 && grid[i - 1][j] == '1') {
                queue.offer((i - 1) * m + j);
                grid[i - 1][j] = '0';
            }
            if (i < n - 1 && grid[i + 1][j] == '1') {
                queue.offer((i + 1) * m + j);
                grid[i + 1][j] = '0';
            }
            if (j > 0 && grid[i][j - 1] == '1') {
                queue.offer((i * m) + j - 1);
                grid[i][j - 1] = '0';
            }
            if (j < m - 1 && grid[i][j + 1] == '1') {
                queue.offer((i * m) + j + 1);
                grid[i][j + 1] = '0';
            }
        }
    }
}