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'

思路：

• 首先能想到的是dfs搜索的方式，因为只要把每一次首次碰到的"1"染色，说明已经来过，那么之后在已经染色的"1"的时候就可以直接忽略
• 染色的过程也可以使用bfs的过程完成

代码

#define bingCJ
#ifdef dfs
//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
    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;
    }
};
#endif

#ifdef bfs
class Solution {
public:
    int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public:
    
    //该函数做的是标记，可以称为染函数，染色标记的意思
    void Dyesig(vector<vector<char>>& grid, vector<vector<bool>>& dye, int x, int y) {
        int n = grid.size();
        int m = grid[0].size();
        queue<pair<int, int>> q;
        q.push({x,y});
        //注意只要是加入了队列就设为真
        dye[x][y] = true;

        while(!q.empty()) {
            pair<int, int> front = q.front();
            q.pop();
            int qx = front.first;
            int qy = front.second;
            //往四个方向扩展
            for(int i = 0; i < 4; i++) {
                int newx = qx + dir[i][0];
                int newy = qy + dir[i][1];
                //判断边界条件
                if(newx < 0 || newx >= n || newy < 0 || newy >= m) {
                    continue;
                }
                //如果在满足边界条件之后该节点没有被染色，那么进行染色操作
                if(!dye[newx][newy] && grid[newx][newy] == '1') {
                    dye[newx][newy] = true;
                    q.push({newx, newy});
                }
            }
        }

    }
    int numIslands(vector<vector<char>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<bool>> Dye(n, vector<bool>(m, false));

        int ans = 0;

        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == '1' && !Dye[i][j]) {
                    ans++;
                    Dyesig(grid, Dye, i, j);
                }
            }
        }
        return ans;
    }
};
#endif

#ifdef bingCJ
class Solution {
public:
    int n, m;
    int totalset;
    int father[90000];
public:
    int find(int x) {
        if(father[x] == x) {
            return x;
        }
        father[x] = find(father[x]);
        return father[x];
    }

    void unionn(int x, int y) {
        int f_x = find(x);
        int f_y = find(y);
        if(f_x != f_y) {
            father[f_x] = f_y;
            totalset--;
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        n = grid.size();
        m = grid[0].size();

       
        //我们总共需要知道一共有多少的陆地，假设他们都是分开的集合
        totalset = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == '1') {
                    totalset++;
                }
                //初始化father为被对应的坐标的大小
                father[i * m + j] = i * m + j;      
            }
        }
        //一共有totalset多的陆地，我们每次合并集合只需要往右下两个方向去合并就可以了
         for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == '1') {
                    if(j + 1 < m && grid[i][j + 1] == '1') {
                        unionn(i * m + j, i * m + j + 1);
                    }
                    if(i + 1 < n && grid[i + 1][j] == '1') {
                        unionn(i * m + j, (i + 1) * m + j);
                    }
                }
            }
        }
        return totalset;

    }
};

#endif