leecode1254.统计封闭岛屿的数目

这道题咋一看很难，但是可以转换，将所有靠近边界的非封闭岛屿都变成海洋后，那么剩下的岛屿就是封闭岛屿，最后的问题其实就是求解连通分量个数 

class Solution {
private:
    void closedIsland(vector<vector<int>>& grid,int x,int y){
        if(x<0||y<0||x>=grid.size()||y>=grid[0].size())
            return;
        else if(grid[x][y]==1)
            return;
        else{
            grid[x][y]=1;
            closedIsland(grid,x,y+1);
            closedIsland(grid,x+1,y);
            closedIsland(grid,x,y-1);
            closedIsland(grid,x-1,y);
        }
    }
public:
    int closedIsland(vector<vector<int>>& grid) {
        int result=0;
        int m=grid.size();
        int n=grid[0].size();
        //将靠近上边界和下边界的非封闭岛变成海洋
        for(int i=0;i<n;i++){
            closedIsland(grid,0,i);
            closedIsland(grid,m-1,i);
        }
        //将左边界和右边界的非封闭岛变成海洋
        for(int i=0;i<m;i++){
            closedIsland(grid,i,0);
            closedIsland(grid,i,n-1);
        }
        //由此问题转化为求岛屿的数目
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                if(grid[i][j]==0){
                    result++;
                    closedIsland(grid,i,j);
                }
        return result;
    }
};