• https://leetcode.cn/classic/problems/number-of-enclaves/description/

• 给你一个大小为 m x n 的二进制矩阵 grid ，其中 0 表示一个海洋单元格、1 表示一个陆地单元格。

• 一次 移动 是指从一个陆地单元格走到另一个相邻（上、下、左、右）的陆地单元格或跨过 grid 的边界。

• 返回网格中 无法 在任意次数的移动中离开网格边界的陆地单元格的数量。

CODE

• 从外向内

class Solution {
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int res = 0;
        int m = grid.size();
        int n = grid[0].size();

        vector<vector<bool>> visit(m,vector<bool>(n,false));

        for(int i=0;i<m;i++){

                dfs(i,0,m,n,grid,visit);

                dfs(i,n-1,m,n,grid,visit);

        }
        for(int i=0;i<n;i++){

                dfs(0,i,m,n,grid,visit);

                dfs(m-1,i,m,n,grid,visit);
        }

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j] ==1)
                    if(!visit[i][j]){
                        res++;
                    }
            }
        }
        return res;

    }

    bool dfs(int i,int j,int m,int n,vector<vector<int>>& grid,vector<vector<bool>>& visit){
        if(i<0 || i>= m) return true;
        if(j<0 || j>= n) return true;
        if(grid[i][j] == 0){return false;}
        if(visit[i][j]) return true;
        visit[i][j] = true ;
        bool up=false,down=false,left=false,right=false;
        {  up  = dfs(i-1,j,m,n,grid,visit);}
        {  down  = dfs(i+1,j,m,n,grid,visit);}
        {  left  = dfs(i,j-1,m,n,grid,visit);}
        {  right  = dfs(i,j+1,m,n,grid,visit);}

        if(up || down || left || right ){
            return true;
        }
        return false;

    }
};

CG

• leetcode200. 岛屿数量

• 错误示例

class Solution {
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int res = 0;
        int m = grid.size();
        int n = grid[0].size();

        vector<vector<bool>> visit(m,vector<bool>(n,false));

        for(int i=0;i<m;i++){
            if(grid[i][0] ==1)
                dfs(i,0,m,n,grid,visit);
            if(grid[i][n-1] ==1)
                dfs(i,n-1,m,n,grid,visit);

        }
        for(int i=0;i<n;i++){
            if(grid[0][i] ==1)
                dfs(0,i,m,n,grid,visit);
            if(grid[n-1][i] ==1)
                dfs(m-1,i,m,n,grid,visit);
        }

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j] ==1)
                    if(!visit[i][j]){
                        res++;
                    }
            }
        }
        return res;

    }

    bool dfs(int i,int j,int m,int n,vector<vector<int>>& grid,vector<vector<bool>>& visit){
        if(i<0 || i>= m) return true;
        if(j<0 || j>= n) return true;
        if(grid[i][j] == 0){return false;}
        if(visit[i][j]) return true;
        visit[i][j] = true ;
        bool up=false,down=false,left=false,right=false;
        {  up  = dfs(i-1,j,m,n,grid,visit);}
        {  down  = dfs(i+1,j,m,n,grid,visit);}
        {  left  = dfs(i,j-1,m,n,grid,visit);}
        {  right  = dfs(i,j+1,m,n,grid,visit);}

        if(up || down || left || right ){
            return true;
        }
        return false;

    }
};

• 超时(从内向外且没有记忆化搜索)

#include <stdio.h>
#include<vector>
#include<memory>
#include<iostream>
using namespace std;
class Solution {
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int res = 0;
        int m = grid.size();
        int n = grid[0].size();

        vector<vector<bool>> visit(n,vector<bool>(m));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j] ==1)
                    if(dfs(i,j,m,n,grid,visit)){
                        res++;
                    }
            }
        }
        return res;

    }

    bool dfs(int i,int j,int m,int n,vector<vector<int>>& grid,vector<vector<bool>>& visit){
        if(i<0 || i>= m) return true;
        if(j<0 || j>= n) return true;
        if(grid[i][j] == 0){return false;}
        visit[i][j] = true ;
        bool up,down,left,right;
        if(i-1>=0)
        if(not visit[i-1][j]){  up  = dfs(i-1,j,m,n,grid,visit);}
        if(i+1<m)
        if(not visit[i+1][j]){  down  = dfs(i+1,j,m,n,grid,visit);}
        if(j-1>=0)
        if(not visit[i][j-1]){  left  = dfs(i,j-1,m,n,grid,visit);}
        if(j+1<n)
        if(not visit[i][j+1]){  right  = dfs(i,j+1,m,n,grid,visit);}
        visit[i][j] = false ;
        if(up || down || left || right ){
            return true;
        }
        return false;

    }
};


int main()
{
    printf("---<");
    unique_ptr<Solution> mysolo = unique_ptr<Solution>(new Solution());
    vector<vector<int>> gird = {{0,0,0,0},{1,0,1,0},{0,1,1,0},{0,0,0,0}};
    int res = mysolo->numEnclaves(gird);
    cout<< res<< endl;
    return 0;
}