刷题51 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

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] 的值为 '0' 或 '1'

题解： dfs

class Solution {
public int numIslands(char[][] grid) {
    int result=0;
    //蔓延算法 dfs
    for(int i=0;i<grid.length;i++){
        for(int j=0;j<grid[0].length;j++){
            if(grid[i][j]=='1'){
                result++;
                spread(grid,i,j);
            }
        }
    }
    return result;
}
public void spread(char[][] grid,int i,int j){
     if(i < 0 || i >= grid.length ||
       j < 0 || j >= grid[0].length || grid[i][j] != '1'){
        return;
    }
    grid[i][j] = '0';
   spread(grid,i-1,j);
   spread(grid,i+1,j);
   spread(grid,i,j-1);
   spread(grid,i,j+1);
    }
}