leetcode 200. 岛屿数量

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
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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);
    }