算法系列——岛屿数量

Posted 2022-01-11 BridgeGeorge

题目

题目链接：力扣

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

解法

岛屿类问题通用解法：力扣

class Solution 
    public int numIslands(char[][] grid) 
        int result=0;
        int row=grid.length;
        int col=grid[0].length;
        for(int i=0;i<row;i++)
            for(int j=0;j<col;j++)
                if(grid[i][j]=='1')
                    result++;
                    dfs(grid,i,j);
                
            
        
        return result;
    

    void dfs(char[][] grid, int r, int c) 
        // 判断 base case
        // 如果坐标 (r, c) 超出了网格范围，直接返回
        if (!inArea(grid, r, c)) 
            return;
        
         // 如果这个格子不是岛屿，直接返回
        if (grid[r][c] != '1') 
            return;
        
        grid[r][c] = '2'; // 将格子标记为「已遍历过」

        // 访问上、下、左、右四个相邻结点
        dfs(grid, r - 1, c);
        dfs(grid, r + 1, c);
        dfs(grid, r, c - 1);
        dfs(grid, r, c + 1);
    

// 判断坐标 (r, c) 是否在网格中
    boolean inArea(char[][] grid, int r, int c) 
        return 0 <= r && r < grid.length 
                && 0 <= c && c < grid[0].length;