力扣 200. 岛屿数量

Posted 2022-03-25 霜序0.2℃

题目

200. 岛屿数量

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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'

解释与代码

三种做法:

• DFS
• BFS
• 并查集

给出DFS的代码

DFS

class Solution 
public:
    int dx[10] = 0, 0, 1, -1;
    int dy[10] = 1, -1, 0, 0;
    void dfs(vector<vector<char>>& grid, int x, int y) 
        if (grid[x][y] == '1') 
            grid[x][y] = '0';
            for (int i=0; i<4; i++) 
                if (x+dx[i]<grid.size() && y+dy[i]<grid[0].size() && x+dx[i]>=0 && y+dy[i]>=0)
                dfs(grid, x+dx[i], y+dy[i]);
            
        
    
    int numIslands(vector<vector<char>>& grid) 
        int ans = 0;
        for (int i=0; i<grid.size(); i++) 
            for (int j=0; j<grid[i].size(); j++) 
                if (grid[i][j] == '1') 
                    dfs(grid, i, j);
                    ans++;
                
            
        
        return ans;
    
;