//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)