LeetCode基础之广度优先搜索 / 深度优先搜索——200. 岛屿数量

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

代码

class Solution {
    public int numIslands(char[][] grid) {
        int count=0;
        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid[0].length;j++){
            // 当前位置为陆地进行BFS(连通的陆地都会更改为水),岛屿数量+1
                if(grid[i][j]=='1'){
                    count++;
                    dfs(grid,i,j);
                }
            }
            
        }
        return count;
    }
    public void dfs(char[][] grid,int i,int j){
    	// 当前位置越界或为水直接返回
        if(i<0||i>=grid.length||j<0||j>=grid[0].length||grid[i][j]=='0'){
            return;
        }
        // 当前位置由陆地改为水
        grid[i][j]='0';
        dfs(grid,i+1,j);
        dfs(grid,i-1,j);
        dfs(grid,i,j+1);
        dfs(grid,i,j-1);
    }
}

标签：陆地,200,优先,int,岛屿,count,dfs,grid,搜索
来源： https://blog.csdn.net/qq_49182770/article/details/123058467