【LeetCode1254】统计封闭岛屿的数目（dfs）

class Solution {
private:
    vector<pair<int,int>>directions{{0,1},{0,-1},{1,0},{-1,0}};
    int num = 0;//岛屿个数
public:
    int closedIsland(vector<vector<int>>& grid) {
        int row=grid.size();//row行数
        int col=grid[0].size();//column列数
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(grid[i][j]==0 && (i == row-1 || j == col-1
                   || i == 0 || j == 0)){
                    //num++;
                    dfs(grid,i,j);
                }
            }
        }
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(grid[i][j]==0){
                    num++;
                    dfs(grid,i,j);
                }
            }
        }
        return num;
    }
    void dfs(vector<vector<int>>&board,int x,int y){
        if(!isarea(board,x,y)){
            return;//如果坐标(x,y)超过网格范围，则直接返回
        }
        if(board[x][y]!=0){
            return;//如果不是岛屿(0)则直接返回
        }
        board[x][y]=2;//将格子标记为已遍历过,重新淹水
        for(auto dir:directions){
            int newx=x+dir.first,newy=y+dir.second;
            if(isarea(board,newx,newy)){//在网格范围内(正常)
                //board[x][y]='X';
                dfs(board,newx,newy);
            }
        }
    }
    bool isarea(vector<vector<int>>&board,int x,int y){//判断点是否在网格内
        if(x>=0 && x<board.size() && 0<=y && y<board[0].size()){
            return true;
        }else{
            return false;
        }
    }  
};