//给你一个由 '1'(陆地)和 '0'(水)组成的的二维网格,请你计算网格中岛屿的数量。 
//
// 岛屿总是被水包围,并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。 
//
// 此外,你可以假设该网格的四条边均被水包围。 
//
// 
//
// 示例 1: 
//
// 
//输入:grid = [
//  ["1","1","1","1","0"],
//  ["1","1","0","1","0"],
//  ["1","1","0","0","0"],
//  ["0","0","0","0","0"]
//]
//输出:1
// 
//
// 示例 2: 
//
// 
//输入:grid = [
//  ["1","1","0","0","0"],
//  ["1","1","0","0","0"],
//  ["0","0","1","0","0"],
//  ["0","0","0","1","1"]
//]
//输出:3
// 
//
// 
//
// 提示: 
//
// 
// m == grid.length 
// n == grid[i].length 
// 1 <= m, n <= 300 
// grid[i][j] 的值为 '0' 或 '1' 
// 
// Related Topics 深度优先搜索 广度优先搜索 并查集 
// 👍 1007 👎 0

/*
* 200 岛屿数量
* 2021-03-04 12:40:11
* @author oxygenbytes
*/ 
#include "leetcode.h" 
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        if (grid.empty() || grid[0].empty()) return 0;
        int m = grid.size(), n = grid[0].size(), res = 0;
        vector<vector<bool>> visited(m, vector<bool>(n));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '0' || visited[i][j]) continue;
                helper(grid, visited, i, j);
                ++res;
            }
        }
        return res;
    }
    int dx[4] = {-1, 0, 1, 0};
    int dy[4] = {0, 1, 0, -1};

    void helper(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
        if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] == '0' || visited[x][y]) return;
        visited[x][y] = true;
        for(int i = 0; i < 4;i++){
            helper(grid, visited, x + dx[i], y + dy[i]);
        }

    }
};
//leetcode submit region end(Prohibit modification and deletion)