鼎湖网站建设公司网站设计方案
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。
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代码实现：
class Solution {public int numIslands(char[][] grid) {int res = 0;for (int r = 0; r < grid.length; r++) {for (int c = 0; c < grid[0].length; c++) {if (grid[r][c] == '1') {dfs(grid, r, c);res++;}}}return res;}private void dfs(char[][] grid, int r, int c) {if (!isInGrid(grid, r, c)) {return;}// if (grid[r][c] == '2') {// return;// }if (grid[r][c] == '0') {return;}grid[r][c] = '0';dfs(grid, r + 1, c);dfs(grid, r - 1, c);dfs(grid, r, c + 1);dfs(grid, r, c - 1);}private boolean isInGrid(char[][] grid, int r, int c) {return r >= 0 && c >= 0 && r < grid.length && c < grid[0].length;}
}
原题链接：力扣
视频讲解：岛屿数量_哔哩哔哩_bilibili