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‘
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
采用回溯，遍历二维数组，每次遇到 1 就把自己和周围的都变成0，并且sum++，最后返回sum即可。
public int numIslands(char[][] grid) { int m = grid.length; int n = grid[0].length; int sum = 0; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { char c = grid[i][j]; if (c == ‘1‘) { sum++; find(grid, i, j); } } } return sum; } private static void find(char[][] grid, int i, int j) { int m = grid.length; int n = grid[0].length; if (i < 0 || i == m || j < 0 || j == n) { return; } if (grid[i][j] == ‘0‘) { return; } grid[i][j] = ‘0‘; find(grid, i - 1, j); find(grid, i + 1, j); find(grid, i, j - 1); find(grid, i, j + 1); }
原文：https://www.cnblogs.com/wangzaiguli/p/14721437.html