岛屿的数量(BFS)

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'

class Solution {
public:int numIslands(vector<vector<char>>& grid) {if (grid.empty()) return 0;int rows = grid.size();int cols = grid[0].size();int islands = 0;int dx[] = {0, 0, -1, 1};int dy[] = {-1, 1, 0, 0};queue<pair<int,int>> q;for(int i = 0; i < rows; i++) {for(int j = 0; j < cols; j++) {  if(grid[i][j] == '1') {      // 发现陆地islands++;               // 岛屿数量加1q.push({i, j});          // 将起点加入队列grid[i][j] = '0';        // 标记为已访问while(!q.empty()) {auto [x, y] = q.front();  // 获取队首坐标q.pop();for(int k = 0; k < 4; k++) {int nx = x + dx[k];   int ny = y + dy[k];  if(nx >= 0 && nx < rows && ny >= 0 && ny < cols && grid[nx][ny] == '1') {q.push({nx, ny}); // 加入队列grid[nx][ny] = '0'; // 标记为已访问}}}}}}return islands;}
};

当遇到 '1'（未访问的陆地）时，说明发现了一个新的岛屿。用 BFS 将所有相连的陆地标记为'0'。

当遍历完后，记录了总的岛屿数量。