【每日一题】【DFS】2021年10月31日--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

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。

 

思路:深度优先搜索DFS

深度优先搜索
class Solution {
    public int numIslands(char[][] grid) {
        int res = 0;
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j ++) {
                if(grid[i][j] == '1') {
                    dfs(grid, i, j);
                    res ++;
                }
            }
        }
        return res;
    }

    public void dfs(char[][] grid, int i, int j) {
        if(i < 0 || i > grid.length - 1 || j < 0 || j > grid[0].length - 1 || grid[i][j] != '1') {
            return;
        }
        grid[i][j] = '2';
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
    }
}

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