岛屿数量

题目

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'

代码

//可以直接在grid上更新但是没有必要
class Solution {
public:
    
    //该函数做的是标记，可以称为染函数，染色标记的意思，并且返回该节点是不是被染色
    void Dyesig(vector<vector<char>>& grid, vector<vector<bool>>& dye, int x, int y) {
        int nr = grid.size();
        int nc = grid[0].size();
        //(x, y)表示的是需要开始做标记的坐标
        dye[x][y] = true;
        //分别表示左右
        if(x < nr - 1 && !dye[x + 1][y] && grid[x + 1][y] == '1'){
            Dyesig(grid, dye, x + 1, y);
        }
        if(y < nc - 1 && !dye[x][y + 1] && grid[x][y + 1] == '1'){
            Dyesig(grid, dye, x, y + 1);
        }
        if(x >= 1 && !dye[x - 1][y] && grid[x - 1][y] == '1') {
            Dyesig(grid, dye, x - 1, y);
        }
        if(y >= 1 && !dye[x][y - 1] && grid[x][y - 1] == '1') {
            Dyesig(grid, dye, x, y - 1);
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        
        int ans = 0;//用来记录岛屿的数量
        int n = grid.size();
        int m = grid[0].size();

        //该二维数组是用来标记是不是染色的，初始化都是没有染色
        vector<vector<bool>> Dye(n, vector<bool>(m, false));

        //深度优先遍历的题目
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == '1' && !Dye[i][j]) {
                    //等于1而且没有被染色
                    ans++;
                    Dyesig(grid, Dye, i , j);
                }
            }
        }
        
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                cout << Dye[i][j] << " ";
            }
            cout << endl;
        }
        return ans;
    }
};