做深度报道的网站在线刷seo

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

思路一：DFS

c++解法

class Solution {
public:void dfs(vector<vector<char>>& grid,int i,int j,int m,int n){if(i<0 || i>=m || j<0 || j>=n || grid[i][j] == '0')return;grid[i][j] = '0';dfs(grid,i + 1,j,m,n);dfs(grid,i - 1,j,m,n);dfs(grid,i,j + 1,m,n);dfs(grid,i,j - 1,m,n);}int numIslands(vector<vector<char>>& grid) {int m = grid.size();int n = grid[0].size();int num = 0;for(int i=0;i<m;i++)for(int j=0;j<n;j++){num += grid[i][j] - '0';dfs(grid,i,j,m,n);}return num;}
};

分析：

本题为岛屿类问题，可用dfs的方式解决，深度搜索将每个遍历过的格子赋值为2即标记为已遍历，后面根据题目需要输出答案，本题是将每个岛屿遍历一遍，当到下一个未遍历的岛屿返回值加一

总结：

本题考察对dfs的应用，利用dfs每当遍历到一个未计数的岛屿则使用dfs将其设为已遍历的岛屿