public int solve (char[][] grid) {
// write code here
int count=0; //返回结果
for(int i=0;i<grid.length;i++){
for(int j=0;j<grid[i].length;j++){
//从第一个开始判断,如果是1进行dfs并count++
if(grid[i][j]=='1'){
dfs(grid,i,j);
count++;
}
}
}
return count;
}
//进行dfs查找
public void dfs(char[][] grid,int i,int j){
//边界限制,以及遇到海洋就终止
if(i<0||i>=grid.length || j<0|| j>=grid[0].length || grid[i][j]=='0') return;
//已经探查过的陆地就使之为“海洋”,避免出现 栈溢出
grid[i][j]='0';
//上下左右
dfs(grid,i+1,j);
dfs(grid,i-1,j);
dfs(grid,i,j+1);
dfs(grid,i,j-1);
}